
Class ___Ll_^ 
Book 
Copyright^ 



COPYRIGHT DEPOSIT 



TEXT BOOKS 

FOB 

ENGINEERS AND STUDENTS. 

BY 

DE YOLSON WOOD, 
Late Professor of Engineering in Stevens Institute of 



A TREATISE ON THE RESISTANCE OF MATERIALS, AND 

AN APPENDIX ON THE PRESERVATION OF TIMBER. 
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This* work originally consisted chiefly of the Lectures delivered by Prof essor Wood 
upon u.is subject to the classes in Civil Engineering in the University of Michigan, but it 
has since been twice revised. Tt is a text-book, and gives a brief sketch of the history of 
the development of the theories connected with the growl h of the subject, and a large 
amount of experimental matter. An English reviewer of the work says : "It is equal in 
grade to Rankine's work." 

A TREATISE ON THE THEORY OF THE CONSTRUCTION 
OF BRIDGES AND ROOFS. Designed as a Text-Book and for 
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The development, of the subject in this work is progressive in its character. It begins 
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eton forms of the triangular trusses, including the Warren girder: and quadrangular forms, 
including the Long, Pratt, Howe, Towne, Whipple, Past, and other forms. The examples 
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THE ELEMENTS OF ANALYTICAL MECHANICS. With numer- 
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PRINCIPLES OF ELEMENTARY MECHANICS. Fully illustrated. 

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A TREATISE ON CIVIL ENGINEERING. By Prof. D. H. Mai: an. 

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THE 



PRINCIPLES 



OF 



ELEMENTARY MECHANICS. 



BY 



DE YOLSONT WOOD, 

Late Professor of Mathematics and Mechanics in the 
Stevens Institute of Technology. 



RE-EDITED WITH EMENDATIONS 
BY 

AKTHUK J. WOOD, 

Professor of Mechanical and Electrical Engineering, 
Delaware College. 



TENTH EDIfTiCN; 
FIRST THOUSAND. 



c •» ■* J 



■ 



NEW YOEK: 

JOHNT WILEY & SONS. 

London: CHAPMAN & HALL, Limited. 

1902. 



THE LIBRARY OF 
CONGRESS, 

Two Copies Recsived 

23 1902 

COPVRWHT ENTRY 

1US&/XX0. No. 

14-1 * %. % 

COPY 3. 



H. 



Copyright, 1877, by De Volson Wood. 
Copyright, 1902, by Abthur J. Wood. 



( . 



PREFACE. 



This work is especially designed to treat of the princi- 
ples of Rational Mechanics, and not to present a system 
of analysis. The analysis employed in the demonstration 
of principles is of an elementary character, the Calculus 
being entirely avoided. A few problems are solved which 
very properly belong to the Calculus, but the solutions 
have been effected by means of the well-known properties 
of certain curves and the principles of elementary geome- 
try. A? examples of this mode of reasoning, reference is 
made to the following problems : The determination of the 
centre of gravity of a circular arc ; The time of vibration 
of a simple pendulum ; and The quantity of flow of a 
liquid through a weir. A few problems are solved which 
involve a knowledge of Conic Sections, but these may be 
omitted, if desired, without detriment to the other portions 
of the work. 

The Articles are as independent of each other as they can 
be practically, and at the same time present the subject in 
a connected manner. This feature will enable the teacher 
the more easily to select particular portions of the work 
when the whole cannot be taken. 

The manner of applying the principles of the subject 



IV PREFACE. 

is shown by means of numerous problems, examples, and 
exercises. The problems are of a general character and 
are accompanied by a full solution. The examples are 
numerical, and are intended to be special applications of 
the formulas and principles contained in the chapter of 
which they form a part. The exercises are a novel feature 
of the work. They are intended particularly to draw out 
and fix in the mind the general principles of the subject. 
The answers to the questions under this head are not 
always explicitly given in the text, but the principle in- 
volved in the answer is sufficiently explained there. Ad- 
ditional questions will doubtless suggest themselves both 
to the teacher and student, and in some cases conditions 
may be added to those given in the exercise. Thus, in 
the 5th Exercise, on page 23, the question may be raised 
whether the weight of the rope is to be considered ; and 
if so, whether the velocity is to be uniform or variable ; 
also, in the latter case, whether the acceleration be increas- 
ing or decreasing. 

The abstract relations which exist between a force and 
the motion which it produces in a body, are considered 
early in the work. I do not consider this order as in 
the least essential, but I have usually presented the sub- 
ject in this way to my classes, regardless of the order 
given in the text-book. No part of abstract mechanics 
possesses a greater interest than this, and the questions 
pertaining to momentum and energy which grow out of 
these relations have provoked a great deal of discussion 
among students in mechanics. In regard to Momentum 



PREFACE. V 

and Vis Viva, much of the difficulty which arises in the 
mind of the student in regard to them would he removed, 
if they are considered, as they really are, mutually inde- 
pendent of each other, having no common unit between 
them, but each having its own peculiar unit. 

The demonstration of the formula for centrifugal force 
may appear to be unnecessarily lengthy, but I trust that 
the student will gain a clearer conception of the mode of 
action of the forces, and be better satisfied with the logic 
of the demonstration, by following the proof here given, 
than by certain of the shorter methods. The demonstra- 
tion by some of the latter methods is defective, although 
the same result is reached by them. 

The principles of energy, which plays such an important 
part in modern physics, have been explained, and the prin- 
ciples of both Kinetic and Potential Energy used in the 
solution of problems. 

De Volsok Wood. 



PREFACE TO TENTH EDITION. 

The general treatment is the same in this as in former 

editions. An index has been added, corrections made, 

and new matter inserted, in order that it may more fully 

meet the requirements of an elementary text-book in 

engineering courses. 

Arthur J. Wood. 
Newark, Del., July, 1902. 



CONTENTS 



[The numbers of the Articles and the leading topic are placed at the head of the page.] 
CHAPTER L 

KINEMATICS. 

Rest. — Motion. — Kinematics. — Rest and motion. — Path. — Velocity.— 
The unit of velocity. — Variable velocity. — Geometrical illustration, 
— General formula. — Circular path. — Angular velocity. — Resolved 
velocities. — Parallelogram of velocities. — Triangle of velocities. — 
Polygon of velocities. — Parallelopipedon of velocities. — Harmonic 
motion. — Periodic motion. — Rotary motion . — Problems. — Exer> 
cises. — Acceleration. —Uniform acceleration. — Formulas. — Initial 
velocity. — Resultant velocity. — Examples. Articles 1-26. 

Pages 1-14. 

CHAPTER II. 

kinetics {commonly called dynamics). 

Blatter. — A body. — Force. — Names for Force. — Measure of force. — 
Weight. — Standard measures. — Force represented by a right line. 
— Point of application. — Inertia. — Molecular motions. — Mechanics. 
— Kinetics. — Statics. — Molar mechanics. — Molecular mechanics. — 
Hydrostatics. — Pneumatics. — Thermodynamics. — Rotation. — Exer- 
cises. — Laws of Motion : — first law — second law — third law. 
— Parallelogram of forces. — Resolved forces. — Components of 
force. — Resultant. — Value of the resultant. — Rectangular compo- 
nents. — Resultant of conspiring forces. — Exercises. — Constant 
force. — Constant acceleration. — Constant moving force. — Illustra- 
tion, — Normal action. — Force of gravity. — Law of universal 
gravitation. — Gravity constant at any place. — Find the accelera- 
tion due to gravity. — Gravity varies with the latitude. — Variable 
weight, causes of. — Atmospheric resistance. — Formulas for fall- 



Vlil CONTENTS. 

ing bodies ; Initial velocity uj or down. — Problems : Spherical 
shell ; Hollow sphere ; Gravity varies as the distance from the cen- 
tre of the earth; Weight. — Mass, measure of. — Unit of mass.— 
Measure of weight. — Density. —Absolute measure of force. - 
Effective force. — Formulas for velocity and space when the accel- 
eration is constant. — Problems. — Examples. Articles 27-90. 

Pages 15-51. 

CHAPTER III. 

WORE.. FRICTION. 

Definition of work ; measure of ; variable. — Motion not work. — Unit of 
work. — Geometrical illustration. — Work independent of time ; 
Time and velocity implied. — Dynamic effect. — Unit of dynamic 
effect. — Useful and prejudicial work. — Work when the force acts 
obliquely to the path. — Work in a moving body (or energy). — Fric- 
tion ; experiments on ; angle of ; laws of ; coefficient of. — Prob- 
lems. — Examples. — Exercises. Articles 91-110 Pages 52-65. 

CHAPTER IV. 

ENERGY. 

Definition ; kinetic; potential. — Heat and work. — Dynamical theory of 
heat. — Mechanical equivalent of heat. Joule's experiment, 
— Transmutation of energy. — Conservation of energy. — Equili- 
brium of energies. — Perpetual motion. — Examples. Articles 111- 
121 Pages 66-77 

CHAPTER V. 

MOMENTUM. 

Definition. — Impulse. — Not a force. — Instantaneous force. — Problems. 
— Impact. — Elasticity; coefficient of. — Elongation of a prismatic 
bar, — Modulus of restitution. — Impact ; non-elastic bodies ; loss of 
velocity ; of perfectly elastic bodies ; Discussion : imperfectly 
elastic bodies; loss of kinetic energy. — Examples. — Exercises.— 
Work, momentum, and energy compared. Articles 122-139. 

Pages 78-93 



CONTENTS. IX 

CHAPTER VI. 

COMPOSITION AND RESOLUTION OF PRESSURES. 

Remark. — Rpsultant pressure; component. — Two equal pressures; 
three equa* pressures having equal angles with each other. — Parallel- 
ogram of pressures. — Direction of resultant; value of. — Triangle 
of pressures ; converse ; proportional to sines. — One force and two 
directions given.— Case of three forces in one plane on a rigid body. 
— Polygon of pressures. — Examples. — Exercises. — Resolution of 
forces. — Rectangular components ; angles a and /3 ; any number of 
forces; resultant of. — Direction of resultant. — Examples. — Forces 
referred to three axes. Articles 140-161 Pages 94-103. 

CHAPTER VII. 

MOMENTS OF FORCES. 

Axis of rotation. — Moment of a force ; force oblique. — Axis of moments. 
— Definitions. — Moment represented by triangle ; sign of ; repre- 
sented by its axis ; composition of. — Moments of three concurrent 
forces in equilibrium. — Unit of moments. — Origin of moments. — 
Arm in terms of two coordinate axes. — Origin of moments. — Par- 
allel forces. — Problems. — Choice of origin. — Problems. — Couples; 
produce rotation ; moment of ; equilibrium of ; resultant of. — Two 
contrary couples on a rigid body. — A force resolved into a couple 
and another force. — If the origin is on the resultant of any number 
of forces, the sum of their moments will be zero. — Three paral- 
lel forces. — A force and couple. — Remark. — Forces in one plane 
if not in equilibrium are equivalent to a single couple, or to a cou- 
ple and force. — If the sine of the moments in reference to three 
points not in a right line are zero, the forces will be in equilibrium. 
—Problems. — Examples. — Exercises. Articles 162-196. 

Pages 104-126. 

CHAPTER VIII. 

PARALLEL forces. 

Definition. — Resultant. — Centre of parallel forces ; to find centre ; re- 
ferred to three axes. — Centre of mass. — Examples. — Exercises, 
Articles 197-203 Pages 127-134 



X CONTENTS. 

CHAPTER IX. 

CENTRE OP GRAVITY OP BODIES. 

Gravity considered as parallel forces ; centre of gravity same as centre of 
parallel forces. — Resultant equals the weight. — Body supported; 
same vertical. — Stable equilibrium ; unstable ; indifferent. — Trial 
methods. — Examples. — Exercises. — Centre of gravity of a part of a 
body ; of several bodies. — Use of coordinate axes. — Examples. — 
Straight line. — Perimeter of a triangle. — Symmetrical figures. — 
Circular arc. — Plane triangle. — Symmetrical areas. — Irregular 
figures. — Zone. — Examples. — Triangular pyramid. — Any pyramid 
or cone. — Spherical sector. — Segment of a sphere. — Examples. — 
Theorems op Papus. — Centre of gravity of a circular arc. — Volume 
computed. — Examples. Articles 204-235 Pages 135-154 

CHAPTER X. 

ENERGY. 

Energy in three states. —Potential Energy. — Equilibrium of a red 
resting against a vertical plane and a curve ; against a pin and 
curve. — Centre of gravity lowest in catenary. — Maximum surface 
generated by the revolution of a line of given length. — Two cylinders 
resting within another. — Equilibrium of a body on an inclined 
plane; pulley; straight lever; bent lever. — Kinetic energy of 
a falling body ; of bodies on two inclined planes ; of one body on an 
inclined plane and the other in a vertical plane. — Movement of 
bodies on a pulley. — Discharge of a fluid through an orifice. — 
Examples. Articles 236-250 Pages 155-169. 

CHAPTER XL 

CONSTRAINED EQUILIBRIUM. 

Definition. — Normal resultant. — Equilibrium on a smooth inclined 
plane; on rough inclined plane. — An eccentric disc. — Equilibrium 
of a beam on two smooth inclined planes. — Equilibrium of a beam 
in a smooth bowl. — Equilibrium of a particle on the arc of a circle ; 
on the arc of a parabola ; on arc of an ellipse ; on arc of a nypei 
bola.— Examples.— Exercises. Articles 251-263 Pages 170-181 



CONTENTS. n 

CHAPTER XIL 

ANALYTICAL METHODS. 

Definitions. — forces not in one plane ; general case ; resolved. — Mo 
ments of resolved forces. — Problems. — Tension of a loaded chord.— 
Weight supported by a string and strut. — Solution of a problem of 
rafters analytically; synthetically. — Examples.— Exercises. Arti- 
cles 264-273 Pages 182-1 93 

CHAPTER XIII. 

STRENGTH OP BABS AND BEAMS. 

Strength of a prismatic piece. --Modulus of tenacity. — Formulas for 
strength. — Law of resistance of beams. — Modulus of rupture.— 
Formula for moment of resistance. — Problems. — Examples. — Arti- 
cles 274-282 Pages 194-201. 

CHAPTER XIV. 

MOTION OF A PARTICLE ON AN INCLINED PLANE. 

Formulas. — Initial velocity. — Problems: motion down the chord of a 
circle ; straight line of quickest descent from a point to a circle ; 
from one circle to another ; from a circle to a line ; from a line to 
a -circle. — Time down a rough plane ; approximate formulas ; equa- 
tions adapted to the movement of cars on an inclined plane. — 
Examples. Articles 283-292 Pages 202-210. 

CHAPTER XV. 

PROJECTILES. 

Definitions ; path and range. — Path, a parabola; velocity at any point 
of ; position of focus; referred to rectangular axes. —The range.— 
Time of flight. — Greatest height. — Greatest range. — Angle of eleva- 
tion for greatest height. — Equal ranges. — Range on an inclined 
plane. — Problems. — Examples. — Exercises. Articles 293-30(5. 

Pages 211-219. 



XU CONTENTS. 

CHAPTER XVL 

CENTRAL FORCES. 

Definition. — Centripetal force; nature of. — The orbit; to find. — Cen- 
trifugal force. — Value of centrifugal force ; in terms of angolac 
velocity ; periodic time. — Centrifugal force of extended masses. — ■ 
Centrifugal force at the equator ; when equal to the weight ; value 
at the moon ; on railroad curves. — Elevation of the outer rail. — 
The conical pendulum. — Examples. — Exercises. Articles 307-323. 

Pages 220-235. 

CHAPTER XVII. 

FORCES WHICH VARY DIRECTLY AS THE DISTANCE PROM THE 
CENTRE OP THE FORCE. 

Remark. — Velocity at any point in the path. — Time of movement. — 
Problems : Simple pendulum ; Compound pendulum ; Length of 
the second's pendulum ; Time lost by clock carried to a height ; 
Movement of a body through the earth ; Vibrations of an elastio 
bar.— Examples. Articles 324-332 Pages 236-249. 

CHAPTER XVIII. 

GENERAL PROPERTIES OF FLUIDS. 

Definition of fluids ; liquids ; uniform bodies. — Forces in the three 
states of matter. — Law of equal pressures. —Normal pressures. — 
Equal transmission of pressures. — Vertical pressures. — Pressure on 
the base of a vessel. — Pressure on the inside of a vessel. — Resolved 
pressures in any direction ; on an immersed body. — Point of appli- 
cation of resultant. — Equilibrium of fluids of different densities. — 
Examples. —Exercises. Articles 333-348 Pages 250-259. 

CHAPTER XIX. 

SPECIFIC GRAVITY. 

Definition. — Specific gravity of a body more dense than water; less 
dense than water ; of a liquid. — Absolute weight of a body. — Speci- 
fic gravity of a soluble body ; of air. — Hydrometers — Definition — 



CONTENTS. XUJ 

Areometer of constant weight. — Nicholson's Hydrometer. — Prob- 
lems ; mechanical combinations ; chemical combinations ; specific 
gravity of a body, the weight in air being given. — Examples. — 
Exercises. Articles 349-362 Pages 260-271. 

CHAPTER XX. 

HYDROSTATICS. 

Compressibility of liquids. — Free surface. — Level surface. — Problems. 
— Examples. — Law of pressure. — Pressure against a rectangle, the 
upper end being in the free surface ; being entirely submerged ; 
against any surface. — Problems. — Examples. — Centre op Pres- 
sure ; of a rectangle ; submerged rectangle ; triangle, base down ; 
base up. — Plane of notation. — Condition of equilibrium of a body ; 
stable equilibrium. — Depth of flotation. — Problems. — Examples. — 
Articles 363-379 Pages 272-288. 

CHAPTER XXI. 

HYDRODYNAMICS. 

Mean velocity. — Permanent flow. — Variable velocity. — Velocity of dia 
charge from an orifice in the base of a vessel ; from an orifice in 
the side. — Head due to the velocity. — Vertical pressure on the free 
surface. — Pressure of the air. — A vertical jet. — Several orifices in 
the side of a vessel. — Oblique jet. — Coefficients of contraction; of 
velocity ; of discharge. — Discharge through a large orifice ; external 
pressure considered. — A weir. — Rectangular notch, quantity of 
flow. — Mean velocity through a weir. — Coefficient of discharge 
through a weir. — Submerged rectangular orifice.— Flow through 
long pipes. — Formulas for circular pipes ; diameter of pipe. — Effect 
of the condition of the pipe. — Flow in rivers and canals ; character 
of the bed of the stream. — Form of free surface in the cross section 
of a stream. — Backwater. — Backwater in rapid streams. — Problems. 
—Examples. —Exercises. Articles 380-409 Pages 289-311. 

CHAPTER XXII. 

GASES AND VAPORS. 

Definition. — Pressure of the atmosphere, Torricellian experiment. — 
Barometer. — Height of a homogeneous atmosphere. — Boyle's (oi 



XlT CONTENTS. 

Mariotte's) law ; law not perfect. — Manometers. — Expansion of gag 
due to temperature. — Coefficient of expansion of air ; of perfect 
gas. — Volume of a gas for given temperature and pressure. — Ther- 
mometers. — Compressing air. — Steam or vapors. — Problems.— 
Weight of a cubic foot of air for any temperature and pressure ; 
Weight of steam ; Spherical air- bubble rising through water. — 
Exhaustion by an air-pump. — To measure elevations with a baro- 
meter. — Flow of a gas into a vacuum. — Examples. — Examples from 
the examination papers for the University of London. Articles 
410-424 Pages 312-335 

APPENDIX L 

TABLE I. — Experiments on friction without unguents 337-340 

" II. — Experiments on the friction of unctuous surfaces. 341 
" III. — Experiments on friction with unguents inter- 
posed 342-344 

" IV. — Specific gravity of bodies 344-347 

Index 353-355 



MEASURES. 



XV 



FRENCH AND ENGLISH MEASURES. 

A DECIMETRE DIVIDED INTO CENTIMETRES AND MILLIMETBES. 



, * 


a 


3 




H 


6 


6 


1 


8 


4 


m 




liulllll 


1 1 1 1 1 1 1 rr 


III ill HI 


TTrrf- 


iiiiin 


Inn 


lllllllll 


lllllllll 


lllllllll 


lllllllll 


Lllllllll' 














1 1 1 I | 1 1 1 1 


l.i 1. L 4 I.I L_L 


1. 1 1 1 1 1 1 1 | 1 1 1 1 1 1 1 1 




l 


2 


»l 


* 



Inches and tenths. 



FRENCH MEASURES IN EQUIVALENT ENGLISH MEASURER 

MEASURES OF LENGTH. 

1 Millimetre = 0" 03937079 inch, or about fa inch. 

1 Centimetre = 0" 3937079 inch, or about 0*4 inch. 

1 Decimetre = 3 937079 inches. 

1 Metre = 39*37079 inches = 3 '28 feet nearly. 

1 Kilometre = 39370 79 incnes. 



MEASURES OP AREA. 

1 sq. millimetre = 0' 00155006 sq. inch. 

1 sq. centimetre = 0" 155000 sq. inch. 

1 sq. decimetre = 15*5006 sq. inches. 

1 sq. metre = 1550 06 sq. inches, or 10.764 sq. feet 



MEASURES OP VOLUME. 

1 cu. centimetre = 0*0610271 cu. inch. 
1 cu. decimetre — 61 0271 cu. inches. 
1 cu. metre = 61027*1 cu. inches. 

The litre (used for liquids) is the same as the cubic deoimetre. 



MEASURES OP WEIGHT. 

1 Milligramme = 0*015432349 grain. 
1 Centigramme = 0* 15432349 grain. 
1 Decigramme = 1 5432349 grains. 
1 Gramme = 15*432349 grains. 

1 Kilogramme = 15432 349 grains, or 2.2 lbs. nearly. 



" Gramme per sq. centimetre 
1 Kilogramme per sq. metre 
1 Kilogramme per sq. millimetre 
1 Kilogramme metre 



TWO UNITS INVOLVED. 

= 2 048098 lbs. per sq. foot 
= 0*2048098 '• 
= 2*048098 " " 

= 7*23314 foot-pounds. 
= 7} foot-pounds nearly. 



1 force de cheval = 75 kilogrammetres per second, or 542£ foot- 
pounds per second nearly. 1 horse-power = 550 loot pounds per 
second = 33,000 loot-pounds per minute. 



XVI 



MEASURES. 



ENGLISH MEASURES IN EQUIVALENT FRENCH MEASURES 



MEASURES OP LENGTH. 

1 inch = 25 39954 millimetres. 
1 foot = 0-304794 metre. 
1 yard = 0' 9143834 metre. 
1 mile = 1-60932 kilometre. 



MEASURES OP AREA. 
1 sq. inch = 645 '137 sq. milli'tres. 
1 sq. foot = 0929 sq. metre. 
I sq. yard = 0- 83609 sq. metre. 
t sq. mile = 2 '59 sq. kilometres. 



MEASURES OP CAPACITY. 

1 pint = 0-5676 litre. 
1 gallon = 4-5410 litres. 
1 bushel = 36-3281 litres. 



MEASURES OP VOLUME. 

1 cu. inch = 16386 6 cu. milli'tres 
1 cu. foot = 0283 cu. metre. 
1 cu. yard = ■ 7645 cu. metre. 



MEASURES OP -WEIGHT. 

1 grain = 0- 064799 gramme. 
1 oz. avoir. — 28*3496 grammes. 
1 lb. avoir. = 0'4535 kilogramme. 
1 ton = 1-01605 tons. 

- 1016-05 kilog. 



TWO UNITS INVOLVED. 



1 lb. per sq. foot = 4 " 88261 kilog. per sq. metre. 
1 lb. per sq. inch = ' 0703 kilog. per sq. centimetre. 
1 foot-pound = 0'1382 kilogrammetre. 







GREEK 


ALPHABET. 






"Letters, 


Names. 


Letters. 


Names. 


Letters. 


Names, 


A a 


Alpha 


I I 


Iota 


P P 


Rho 


B/3 


Beta 


K K 


Kappa 


2 <T s 


Sigma 


r r 


Gamma 


A X 


Lambda 


T T 


Tau 


A 8 


Delta 


M fl 


Mu 


Y v 


Upsilon 


E e 


Epsilon 


N v 


Nu 


$ (f> 


Phi 


z ? 


Zeta 


3 £ 


Xi 


x X 


Chi 


H V 


Eta 


O o 


Omicron 


¥ ^ 


Psi 


000 


Theta 


n w 


Pi 


Q a) 


Omega 



ELEMENTARY MECHANICS. 



CHAPTER I. 

KINEMATICS. 

Preliminary Notions. 

1. Rest. — Bodies are said to be at rest when they 
remain in the same place in reference to surrounding 
objects. 

2. Motion. — A body is said to be in motion relative 
to another body when it changes its position with respect 
to that body. We may mention here two different 
kinds of motion, each complying with the above defini- 
tion: (1) motion of translation, when the change of 
position is that of the body as a whole with reference to 
other bodies external to it ; in this case all points of the 
body describe paths equal in magnitude and parallel in 
direction. (2) If a point moves so that its direction is 
continually changing, the path is a curve. The point 
will have curvilinear motion. 

3. Kinematics is the science of motion. — It does not 
consider the cause of the motion, but determines its 
measure, and the relations between different motions. 

4. Rest and motion are relative terms. — A body may 
be at rest in reference to some objects, and in motion in 
reference to others. Thus, a person sitting on the deck 
of a ship may be at rest in reference to the objects on 
the ship, while he is moving with the ship over the water. 
But if he should run toward the stern at the same rate 
that the vessel is advancing, he will appear to be at rest 
in reference to objects on the shore, and moving in refer- 
ence to the objects on deck. Objects at rest on the eartn 
are moving through space with the earth at the rate of 
nearly 19 miles a second. 

1 



2 KINEMATICS. [5,6] 

The motion of one body in reference tc another also ir, 
motion is called relative; but in reference to a fixed 
object it is called absolute or actual. 

5. The path of a body is the li?ie traced by its central 
point. — If all the points of a body move in parallel lines, 
any one of the lines may be taken as the path. Unless 
otherwise stated, we will assume that the body is reduced 
to a mere particle. The path is also called the space over 
which a body moves. 

Velocity. 

6. Velocity is rate of motion. — When a body passes 
over equal successive portions of space in equal times, its 
rate is uniform. In all other cases it is variable. Thus, 

if a body moves uniformly 

1 3 3 4f c 

■A' • * >— — -*3 from A to B in four seconds, 

the spaces passed over each 
second will be one-fourth of AB, or A-l = 1-2 = 2-3, etc« 
and any one of these spaces is the velocity. 

When the motion is uniform, the velocity is the space 
passed over in a unit of time, and the velocity is said to 
be constant. 

If s = the space passed over uniformly, 
t = the time of the movement, and 
v = the velocity, 
then we have, according to the definition, 

•=7; • • a) 

from which we find, 

8 = vt; . . (2) 

and 

* = J. • • • (3) 



[7-9.] VELOCITY. 3 

7. The unit of velocity is understciod to be one foot 
per second, unless otherwise stated. If other units are 
given, their equivalent value may be found in feet per 
second. 

8. Variable Velocity is that in which the rate of 
motion is constantly changing. — The true measure in this 
case cannot be the space passed over during any single 
second, but it is the space which would he passed over 
during a second if the body moved uniformly at the rate 
which it had at the instant considered. 

We are familiar with this fact. We say that a train of 
cars moved at the rate of, say, forty miles per hour, when 
it may have moved at that rate for an instant only, and in 
coming to rest it may have moved at all conceivable rates 
less than forty miles per hour. 

9. Geometrical Illustration. — Variable velocity may 
be represented by a figure. Thus, in Fig. 2, let AB 
represent the time, say four seconds. 

Divide it into four equal parts, each 

of which will represent one second. 

At the several points of division erect 

ordinates, and make them proportional 

to the velocities corresponding to those 

times ; the ordinate al representing the 

velocity at the end of one second ; b2 at 

the end of two seconds, and so on. A Fl °* 2, 

curve AabcO 'may be drawn through the extremities of these 

ordinates, such that the ordinate at any point will represent 

the velocity corresponding to that time. If a line ad be 

drawn through a parallel to AB, the number of square 

units in lad% will also represent the velocity ; for the side 

1-2 of the rectangle being unity, there will be the same 

number of square units in the rectangle that there are 



4 KINEMATICS. [10,13.] 

linear units in al, and similarly for any other part of the 
figure. 

The area of the figure ABC will represent the number 
of units of space passed over by the body in four 
seconds. 

10. The general computations for variable velocity 
belong to higher mathematics ; but we are enabled to 
treat of some cases in an elementary manner, as will be 
shown hereafter. 

The velocity may be found in practical cases with suffi- 
cient accuracy by finding the actual space passed over by 
a body in a \erj short space of time, and considering the 
motion as uniform during that time. 

If At (read element of time, or simply delta t) be the 
element of time, and 

Js = the corresponding space, 
then, according to the definition, we have 

~As 



At 

11. The path may be the arc of a circle, or any other 

curved line, in which case the space will be the length of 

the developed line. If AB be the 

arc of a circle passed over in time t, 

we have v = AB -j- t, as before. 

12. Angular Velocity is the rate 
of angular movement. It is measured 
by the circular arc having unity for its 
radius, which would be generated by 
the extremity of the radius if it turned 
about the centre at the same rate as that of the body. 11 
AB = v } and Oa = 1, then ab = angular velocity. 





f 13.} VELOCITY. 5 

Or, if A B = 8 — the space passed over uniformly by 

the body, 
t = the time, 

r = OA = the radius of the arc AB, and 
w = the angular velocity, 
then 

o 

— = the velocity along AB, and 
t 

s 1) 

w = - — ■. **. w = -, when the motion is uniform. 

13. Resolved Velocities. — Suppose that a body moves 
uniformly from A to B. At the extremities of the line 
draw two lines, AG and BO, 
forming a right angle at C. 
Then, if a body moves from A 
to C in the same time that one 
moves from A to B, the velocity 
of the former will equal the latter into cos. BAG 
Or, if 

v = the velocity along AB, 
v x = the velocity along A G, and 
= angle BAG; 
then, 

Vi = v cos. p. . . . (1) 
Similarly, 

v 2 = v sin. p. . . . (2) 

when v 2 is the uniform velocity along BO. 

Suppose that a body is pushed in a due southerly direc- 
tion, parallel to BO, and at the same time in a westerly 
direction parallel to OA ; if the velocities in these direc- 
tions are uniform and proportional to BO and OA re- 
spectively, the resultant motion will be along a line paral 
lei to BA, and the velocity will be proportional to BA. 



6 KINEMATICS. [14,15.] 

The velocities represented by BG and GA are called 
component velocities, and that by BA is called the result- 
ant velocity. 

14. Parallelogram of Velocities. — The component 
velocities AG and BG may make any angle with each 
other. Thus, if in Fig. 5 

v = the velocity along AB, 
v t = the velocity along A G, 
v % = the velocity along AD, 
P = the angle Z> A G; 

then, from plane trigonometry, we have 

AB = ^/AG 2 + AD* + 2AGAB cos. BAG; 
or, 

v = tf v *+ v * + 2^ cos. 0. 

If the angle BA G is obtuse, cos. j3 will be negative. If 

BA G is a right angle, we have 
cos. 90° = 0, and 




v = W + vj. 

This principle may be stated 
as follows : 

If two velocities are repre- 
sented in magnitude and direction by the adjacent sides 
of a parallelogram, the resultant velocity will be repre- 
sented in magnitude and direction by that diagonal of the 
parallelogram which lies between these sides. 

15. Triangle of Velocities. — If two concurrent veloci- 
ties be represented in magnitude and direction by two sides 
of a triangle taken in their order, the resultant velocity will 
be represented in magnitude and direction by the third 
side. 



[16-18 j 



VELOCITY. 




Thus, in Fig. 5, if A C represent the velocity v x in the 
direction from A towards C, and CB represent the velo- 
city v. z , acting from C towards B, then will AB represent 
the resultant velocity. This is evident from the preced- 
ing article. 

16. Polygon of Velocities. — If several velocities, act- 
ing successively, carry a body around a polygon, they will 
produce rest if they all act at 
the same time. And, if several 
velocities be represented in mag- 
nitude and direction by the 
successive sides of a polygon 
taken in their order, when they 
act all at the same time their 
resultant velocity will be repre- 
sented by the closing side of the polygon. 

This is proved by means of the preceding article, by 
compounding two velocities, then their resultant and a 
third velocity, and so on to the last. 

17. Parallelopipedon of Velocities. — If three veloci- 
ties not in one plane be represented by the three adjacent 
edges of a parallelopipedon, the result- 
ant velocity will be represented by that 
diagonal which passes through the 
common point of the three edges. 

This may also be proved by the 
Triangle of Yelocities. The resultant 
velocity of a body which has a south- 
ward, westward, and downward mo- 
tion, will be southwesterly and downward. 

18. Harmonic Motion. — If a body moves at a uniform 
rate around the circumference of a circle, the foot of the 
perpendicular from the body upon the diameter will 




Fig. 7. 




Fig. 8. 



8 KINEMATICS. [19-21.1 

appear to move to and fro along the diameter with a varia- 
ble velocity. Thus, in Fig. 8, if the point move uniformly 
around the circumference A OB, the foot of the perpen- 
dicular will move from A towards B^ 
thence from B towards A, and so on. 
The motion of the point is said to 
be harmonic, for the law of its move- 
ment is similar to that of a musical 
string, or a tuning fork. 

19. Periodic Motion is that in 

ivhich the motion repeats itself. — Thus, 

in Fig. 8, the point that moves to and fro along the diam 

eter AB has a periodic motion. A pendulum, as it 

vibrates to and fro, is another example. 

20. Rotary Motion is motion about an axis. It is 
measured by its angular velocity. See Article 12. 

The point about which it moves may also have a pro- 
gressive velocity. Thus, the wheels of a carriage have a 
rotary motion about the axles, while the axles have a 
progressive movement. The moon revolves about the 
earth as a centre, and the earth not only revolves on its 
axis, but also revolves around the sun. 

21. Problems. — 1. If the circumference of the earth 
at the equator is 25,000 miles, what is the velocity in feet 
jper second of a point on the equator? 

The earth turns on its axis once in twenty-four hours, 
and there are 5,280 feet in a mile ; hence, the velocity in 
feet per second is 

25000 x 5280 1JCO/r . 
24 x 60 x 60 



2. Required the angular velocity of the ea/rth. 



[21.] VELOCITY. 9 

The circumference of a circle, whose radius is unity 
is 2t = 6.28 + ; hence, the angular velocity per hour is 

. = *» 0J8+. 

24 



The angular velocity per hour, in terms of degrees is 
i 
which per minute is 
w 
and per second is 



360° 



-60 "4 ~ 15 ' 



— §=*" 



EXERCISES. 

1. If one body moves at the rate of 10 miles per hour, and another 

body, starting from the same place, moves in an opposite direction 
at the rate of 15 miles per hour, both moving uniformly, find the 
distance between them at the end of 10 minutes. 

2. Which moves faster : a body moving 6 feet per second or one moving 

100 yards per minute ? 

3. A railway train travels 90 miles in two hours ; find the velocity in 

feet per second. 

4. Two bodies start from the same place at the same time, and move 

Uniformly at right angles with one another, one at the rate of 8 feet 
per second, the other at the rate of 15 feet per second ; find the 
distance between them at the end of one second ; also at the end 
of n seconds. 

5. If a train moves uniformly at the rate of 40 miles per hour, how 

many seconds will it take to go 400 feet ? 

6. If the fly-wheel of an engine revolves 200 times per minute, what 

will be its angular velocity per second ? 

7. If a person aims to row directly across a stream at the rate of 3 miles 

per hour, while the stream carries him downward at the rate of 
2 miles per hour, at what rate will the boat move ? 
1* 



10 KINEMATICS. [22,23.] 

8. If a wheel is rolled directly across the deck of a ship at the rate of 

15 feet per second, while the ship is moving 10 miles per hour, 
find the velocity of the wheel in space. 

9. If the velocity is one metre per hour, find the velocity in feet pei 

second. 

(One metre is 3.28 feet nearly.) 

10. Two bodies start from the same plaoe at the same time, and move 

in paths which are inclined 60 degrees to each other, one moving 
at the rate of 5 feet per second and the other at the rate of 10 
feet per second ; required the distance between them at the end 
of two seconds. 



Acceleration. 

22. Acceleration is the rate of change of the velocity. 
— If the rate of motion be uniform, the velocity is con- 
stant, and there will be no acceleration. If the velocity 
constantly increases, the rate of change is called a positive 
acceleration, but if it constantly decreases it is called 
negative. The rate of change may be uniform or con- 
stantly changing. If the acceleration is uniform, it is 
measured by the amount hy which the velocity is increased 
in a unit of time ; and if it he variable, it is measured by 
the amount by which the velocity would be increased in a 
unit of time if its rate of increase continued the same as 
at the instant considered. The unit of time is usually 
one second. If the velocity is decreasing, the same defi- 
nition applies by substituting the word decreased for 
increased. 

23. Uniform Acceleration may be represented by a 
triangle. ' Thus, in Fig. 9, let AB represent the time — say 
four seconds — and BO the velocity at the end of four 
seconds, the body having started from rest. Divide AB 
into four equal parts at the points b, g, h, and draw the 
horizontal lines be, ge, M, to the line drawn from A to O; 



[24.] 



ACCELERATION. 



11 




then will be represent the velocity at the end of the first 
second, ge, at the end of the second second, and so on. 
Draw the vertical line cd; then will de represent the 
increase of the velocity during the 
second second, and hence is the accel- 
eration. Similarly fh will represent 
the acceleration during the third 
second, and iC that during the fourth. 
But iC=fh = de = bc; or the velo- 
city increases uniformly and the accel- 
eration is constant, and the velocity at 
the end of the first second equals the 
acceleration. 

The space passed over during the second second will be 
represented by the trapezoid bceg, which is 3 times the 
triangle Abe, The space gehh is 5 times the triangle Abe. 
Hence, generally, if the times are represented by the 
natural numbers 

1, 2, 3, 4, etc., the spaces will be 
1, 3, 5, 7, etc. 

When the acceleration is uniform, the spaces described 
may be laid off on a straight line, as is shown by the line 
AC, Fig. 11. 
24. Formulas for uniform acceleration. 
Let 

f= the uniform acceleration, 
t = the time, 

8 = the space passed over in the time t, 
v = the velocity at the end of the time t ; 



then, in Fig. 9, we have 



12 



KINEMATICS. 



F36.J 



or 



Al\lc\\AB\BG\ 
l:f::t:v; 



/. v =ft. . . 
Area ABC =%BCxAB; 
s = %vt ; . . 
and by eliminating t we have 



Also, 
or 



s = 



¥' 



(i) 

(9) 
(3) 



Eliminating v between equations (1) and (2) gives 

s = iff. ... (4) 

When any two of the quantities f\ v, t, s are given, the 
other two may be found from the preceding equations. 

25. Initial Velocity.— The velocity which a body has 
at the instant t begins to be reckoned, is called initial 

velocity. This may be illustrated 
by Fig. 10, in which AB repre- 
sents the time, DA the initial 
velocity, EC the final velocity, 
and DEC A the space. It thus 
appears that the final velocity will 
be that due to the acceleration 
plus the initial velocity ; and the 
space will be that due to a uniform movement equal to 
the initial velocity, plus that due to the acceleration ; 
hence, if 

v = the initial velocity, 

tVo = ABED = the space due to the ini 




Fig. 10. 



then 



rse.j 



ACCELERATION. 



13 



tial velocity ; and using the same notation as in the pre- 
ceding article, we have for this case 



s = tv + 

s = tv + \ff. 

If the acceleration is decreasing, we have 

v = v Q —ft, 
s = tv — iff. 



(1) 
(2) 

(3) 



(5) 



26. The Resultant of Variable and Constant Velo- 
cities. — In Fig. 11, let the horizontal velocity be constant, 
and the vertical motion be uniformly accelerated. If Aa 
is the space passed over in a horizon- 
tal direction during the first second, 
arid Ad, the vertical space during the 
same time, then will the body be at 
the intersection of the lines drawn 
respectively through a aud d; the 
former, vertical and the latter, hori- 
zontal. In a similar way its position 
may be found at the end of any given 
time. 

The locus of these points will be 
the path of the body, and will be a parabola. 

In a similar way the path may be found, when both 
velocities are variable and acting at any angle with each 
other. 




Fig. 11. 



EXAMPLES. 

If a body starts from rest and has a uniform accelera- 
tion of 5 feet per second, how far will it move in 10 
seconds, and what velocity will it acquire ? 



14 KINEMATICS. [26.] 

2. A body starts from rest and acquires a velocity of 

100 feet in 4 seconds ; required the acceleration and 
the space passed over during the first second. 

3. If a body starts from rest with a uniform acceleration 

of 32 feet per second, how far will it move during 
the 4th second ? 
h A body has an initial velocity of 50 feet per second, 
and has a velocity of 100 feet at the end of 8 seconds ; 
required the acceleration in feet per second. 

5. If the acceleration is uniformly 32 feet per second; 

required the time necessary to pass over 200 feet, 
the body starting from rest. 

6. If the acceleration is 20 metres in two minutes, what 

will be the acceleration in feet per second % 

7. If the acceleration is 32£ feet per second, what will be 

the value in metres per second % 



CHAPTEE EL 

KINETICS t 

(Commonly called Dynamics,) 

27. Matter is the substance of which bodies are com- 
posed. — In its grosser forms we gain a knowledge of it 
by common experience. It is difficult, if not impossible, 
to define it so that a person who is not already familiar 
with it will gain a correct notion of it. It has certain 
properties, such as extension, divisibility, porosity, elas- 
ticity, etc., which it is the province of works on physics to 
investigate. For the purposes of Mechanical Science, it 
may be defined as that upon which force acts. But, when 
we consider the effect of forces upon bodies, it is necessary 
to know, or to assume, certain properties, especially the 
compressibility and elasticity of matter. 

28. A Body is a finite portion of matter. — An atom is 
an indivisible portion of matter. It is an ideal thing, 
since we know nothing of its essential nature, although it 
has been a subject of much speculation. It is assumed 
that a body may be divided and subdivided, until parts 
might be reached which, from their constitution, cannot 
be again divided. A molecule is the smallest portion of a 
body which can exist without changing its nature. It is 
composed of two or more atoms. Thus, a molecule of 
water is composed of two atoms of hydrogen and one atom 
of oxygen, and if they be separated the result is no longer 
water but two distinct substances. 

A particle is a very small body, or a small portion of a 



16 KINETICS. [ 29, 30. i 

body. It may be composed of several molecules. It has 
no reference to the constitution of the body. In mechan- 
ics it is considered as a material point. 

According to the speculations of Sir W. Thomson, 
Maxwell, Tait and others, the diameter of a molecule ex 

ceeds mm* of an iuch ' and is less than sopkoob of an inch - 
In order to give an idea of the minuteness of a mole- 
cule, Sir W. Thomson states that if a body of water of the 
size of an ordinary pea be expanded to the size of the earth, 
all the molecules expanding in the same ratio, the mole- 
cules would be between the sizes of fine shot and a cricket 
ball. 

A rigid body is a body whose particles do not change 
their positions with respect to one another. 

Force. 

29. Force is that which changes or tends to change the 
state of a body in reference to rest or motion. — It is a 
cause, the essential nature of which we are ignorant of. 
We deal only with the laivs of its action. These laws are 
determined from observation combined with certain com- 
putations. All forces do not act according to the same 
laws. Thus, it has been found that the force of gravity 
varies inversely as the square of the distance between the 
centres of attracting spheres ; in elastic bodies the force 
resisting distortion varies directly as the amount of displace- 
ment ; and other forces may vary according to other laws. 

30. Forces are called by different names, according to 
the results produced or the phenomena presented. Thus 
we speak of attraction, repulsion, cohesion, friction, 
moving forces, accelerating forces, resisting forces, con- 
stant forces, variable forces, muscular forces, vital forces, 
etc. ; but they are all alike in the essential quality, that 
they are equivalent to a, pull or &jpush. 

31. Measure of Force. — We shall assume that the 



[31-33.] FOEOE. 1? 

standard pound Avoirdupois is the measure of a, unit of 
force, and hence, that any force is a certain number of 
pounds. We are familiar with the fact that forces are 
measured by pounds. Thus, if a spring balance of suffi- 
cient strength is placed between a locomotive and a train 
of cars, it will indicate the pulling force of the locomotive, 
whether the train remains at rest or is in motion. A push 
can be measured in the same manner. 

Forces may also be measured by considering their effect 
in producing motion in a free body, as will be shown in 
Article 86. 

32. Weight is a measure of the attractive force of 
gravity upon a body. — Weight varies directly as gravity. 
It has been found that the same body will weigh more in 
some places than others. It will weigh a trifle less on the 
top of a high mountain than in a deep valley, and if it 
could be placed where there was no attraction it would 
weigh nothing. It is, therefore, necessary to designate 
some place where a body, used as a standard, shall be 
weighed in order to fix the standard pound. 

33. Standard Measures. — The standards for weights 
and measures are established by legal enactments. The 
British standard for one pound Avoirdupois is the weight 
of a certain piece of platinum kept in the Exchequer Office 
in London. 

The British standard yard is the distance between two 
points on a certain metal rod, kept in the Tower of Lon- 
don, when the temperature of the whole bar is 60° F. 
(=15°.5C). The foot is declared to be one-third of the yard. 

The United States standards were copied from the 
British standards; but it has since been found that, on 
account of errors in measurement, the British yard is a 
trifle shorter than the American. 



18 KINETICS. [33.] 

The French Tnv&re is the distance between two points on 
a certain bar kept for the purpose, and is nearly i 0<0 oo,ooo of 
the length of a meridian measured from the equator to the 
north pole. 

The relation between these standards and certain definite 
quantities furnished by nature, has been determined with 
great care. For instance, after Capt. Kater determined 
very accurately the length of a pendulum which would 
vibrate once a second at London, compared with the stand- 
ard then in use, it was declared by Parliament (5 Geo. IV.), 
that " the yard shall contain 36 parts of the 39.1393 parts 
into which that pendulum may be divided which vibrates 
seconds of mean time in the latitude of London, in vacuo, 
at the level of the sea, at temperature 62° F." After the 
standard was destroyed by fire, the commissioners who 
examined the subject reported that several reductions of 
the pendulum experiments were doubtful or erroneous, 
and, accordingly, the distance between the marks on a 
metallic bar was adopted as the standard, and the above 
ratio was discarded. 

Similarly, the metre was originally declared to be 
10,000,000 of the arc of the meridian measured from the equa- 
tor to the north pole, and the French government expended 
large sums of money in determining this distance. Cer- 
tain arcs of the meridian were measured with great care, 
and from them the distance was computed. Afterwards 
it was ascertained that there had been errors in the deter* 
mination, and that the distance was not the same on all 
meridians, for the equator is not an exact circle ; but the 
length of the metre was not changed, and hence the 
declared ratio became obsolete. 

Attempts have been made to establish a legal pound by 
declaring that a cubic foot of water at its maximum 



[34,35.] FORCE. 10 

density shall weigh a given amount (nearly 63i pounds); 
but there is a question in regard to the temperature of the 
water at the maximum density, and different experimenters 
do not agree as to the weight. Hence, practically and 
actually, the weight of a certain piece of metal remains 
as the standard. (See The Metric System, by F. A. P. 
Barnard, LL.D., New York, 1872.) The Avoirdupois 
pound is 7,000 grains. For the relation between the 
English and French measures see table on pp. xvi and xvii. 
The French metre is equivalent to 39.37079 British inches, 
or 39.368505 American inches. 

The unit of time in common use is a second of mean 
solar time. The time from any meridian passage of the 
sun to the following one is called a solar day, or simply a 
day. These days are of unequal length, caused by the 
variable motion of the earth along its orbit, while the 
time of the revolution of the earth on its axis is con- 
stant. Therefore, the average length of the day for an 
entire year is used, and called 24 hours, from which the 
minute and second are easily found. The second is the 
86,400th part of a msan solar day. 

34. A force may be represented by a straight line. 
Thus, in Fig. 12, the magnitude of the force may be 
represented by the length of the line 

AjB, the point of application, by the ^ — > & 

point A, and its direction of action, by FlG ' 12 ' 

the arrow-head, indicating that it acts from A towards B. 

35. The point of application of a force may be con- 
sidered as at any point in its line of action. Thus, if a 
body is pushed with a rod, the point of application of the 
force may be considered as at any point along the line of 
the rod ; and, if the rod were prolonged through, and 
beyond the body, the point of application may be consid 



20 KINETICS, [S6. 1 

ered as at any pvint on the prolonged part. The effect 
upon the body will be the same, whether we consider the 
point of application of the force at one point or another 
of the rod. 

Inertia, 

36. The Inertia of matter means its jpassiveness / that 
is, its inability to change its own condition in regard to 
rest or motion. If a body be at rest, it has no power to put 
itself in motion, or, if in motion, it has no power of itself 
to change its rate of motion. It implies that every change 
of motion is due to an external cause. 

A body not constrained by other bodies, or in other 
words, perfectly free to move, is perfectly sensitive to 
the action of a force, so that the smallest force would 
move the largest body. The moon, earth, and other 
planets are in this condition, and each yields constantly to 
the action of all the others upon it. It is said, and with 
good reason, that every step which a man takes upon the 
earth changes the centre of gravity of the earth. 

Inertia is not a force. But on account of the perfect 
passiveness of matter, if a force act upon a free body, the 
effect of the force will be stored in the body ; and the body, 
in being brought to rest by a resistance, will produce the 
same effect as that which was stored in it. 

The exact relations which exist between the motion of 
bodies and the force which produces it, were established 
only after a long series of observations, experiments and 
deductions. It is difficult to prove them by direct experi- 
ments, for it is difficult to realize the condition of a per- 
fectly free body. Every body is liable to be resisted by 
friction, or the air, or other medium ; and if the body be 
placed in a vacuum, the range of its motion will be limited 



[37-43.] MOLECULAR MOTIONS. 21 

If a body be thrown into the air, it not only meets with 
the resistance of the air, but will be constantly under the 
action of the force of gravity. Delicate experiments have, 
however, confirmed all the fundamental principles of mo- 
tion. Their truth is also shown from the fact that all prob- 
lems pertaining to the motion of bodies, not only on the 
earth but in the solar system, solved in accordance with 
these principles, give results which agree with the results 
of observation. The times and places of eclipses are pre- 
dicted, and the positions of the planets are foretold, by 
means of formulas which grow out of these fundamental 
princi pies. No truths in science have become more firmly 
established. 



FUNDAMENTAL DEFINITIONS. 

37. Mechanics is the science which treats of the action 
of forces. — It investigates the laws which govern the 
action of forces ; the conditions of the equilibrium of 
bodies ; the motion of bodies under the action of forces ; 
the flow of liquids and gases ; and the movement of the 
particles of bodies. 

38. Kinetics treats of the movement of bodies under 
the action of forces. — This branch of the subject has 
usually been called Dynamics, a term which more properly 
pertains to the abstract doctrine of forces, and hence, as 
such, would include a portion of both Kinetics and Statics. 

39. Statics treats of the conditions of the equilibrium 
cf solids. 

40. Molar mechanics treats of the action of forces upon 
solids. 

41. Molecular mechanics treats of the movement of 
the particles of a body. 

42. Hydrostatics treats of the equilibrium of liquids, 

43. Hydrodynamics treats of the movement of liquids.. 



22 KINETICS. [44-47. J 

44. Pneumatics treats of the laws of pressure and 
movement of air and other gaseous bodies. 

45. Thermodynamics treats of the mechanical prop- 
erties of heat. 

46. Rotation of Bodies. — A force acting upon a body 
may produce both translation and rotation at the same 
time. Thus, a boy strikes a stick with a club, and if he 
does not strike it directly opposite the centre it will rotate 
at the same time that it moves forward. The solution of 
problems involving the rotation of bodies is generally 
more difficult than those involving translation only. If all 
the points of a body move in parallel straight lines, the 
motion is that of simple translation, and will be substan- 
tially the same as if we consider the body reduced to a 
mere particle. Hence, in discussing the subject of trans- 
lation it is common to speak of the body as a particle. 

Molecular Motions. 

47. Molecular Motions. — We have thus far spoken of 
the motion of bodies, but a close examination shows that 
the particles of a body may have a motion in reference to 
each other. Thus, when a tuning-fork or a bell is struck 
the particles are put into a rapid vibratory motion, 
producing sound, which is transmitted to the ear by the 
vibrations of the air. Heat expands bodies, an effect 
which must be caused by the separation of the particles of 
the body. It is believed that the molecules or atoms of 
every solid are made to vibrate when it is struck. 

EXERCISES. 

1. If a body be suspended by a very long, fine string, how much foroa 

will be required to move it sideways ? 

2. If a ship could float on water without any resistance from the water 

or air, bow much force would be required to move it ? If it were 



[48-60.] NEWTON'S LAWS OF MOTION. 23 

moving, how much force would be required to deflect it from its 
path? 

3. Why does not every body move when acted upon by force ? 

4. If four spring balances are connected end to end, and a man pulls 

with a force of 100 pounds at one end of the four, what will be the 
force exerted at the other end, and what will each balance indicate ? 
6. If a constant pull of 500 pounds is exerted at one end of a rope, 
which is attached to a boat at the other end, will the force exerted 
at the other end be 500 pounds when the boat is in motion ? 

6. Suppose that two heavy sleds are placed on perfectly smooth ice and 

connected by a flexible cord ; if a boy draws them by pulling with 
a constant force of 10 pounds on one of them by means of a rope 
or otherwise, so as to pull both sleds in the same direction, will 
there be a force of 10 pounds exerted on the other one ? That is, 
will the tension on the connecting cord be 10 pounds ? 

(Perfectly smooth is intended to imply that there is no resistance 
between the sleds and ice.) 

7. If, in the preceding exercise, the boy ceases to pull, but all the other 

conditions remain the same, what will be the tension upon the 
connecting cord ? 

Newton's Three Laws of Motion. 

48. Sir Isaac Newton expressed the fundamental prin- 
ciples of the relations of force to the movement of bodies 
in the form of three laws or axioms. These principles 
have already been given in the preceding articles, but 
these laws are referred to so frequently, and express so 
clearly and correctly the fundamental principles of the 
motion of bodies, that we cannot do better than present 
them in this place. 

49. First Law. — Every body continues in a state of 
rest or of uniform motion in a straight line, unless acted 
upon by a force which compels a change. 

50. Second Law. — Change of motion is in proportion 
to the acting force, and takes place in the direction of the 
straight line in which the force acts. 

Observe that the resultant motion is not necessarily in 



24 KINETICS, \ 51, 52. j 

the direction of action of the deflecting force, b .t it im- 
plies that the departure from the line of motion will be 
proportional to the deflecting force. To illustrate: if a 
body be moving due south, and a force be applied which 
acts in a due easterly direction, like the wind, for instance, 
then will the easterly force cause it to move east of the 
meridian in which it originally moved, and the amount of 
this departure will be proportional to the easterly force, 
and will be independent of the velocity in the southerly 
direction. 

51. Third Law. — To every action there is always 
ojijwsed an equal and opposite reaction. 

That is, when two bodies press one against the other, 
the action upon one is exactly equal but opposite to that 
upon the other. If one presses a stone with his finger, the 
foroe acts equally against the stone ami the finger. 

The term motion in these laws means more than velo- 
city ; it really means momentum, a term which will be 
defined hereafter. 

Resolved Forces. 

52. Parallelogram of Forces. — If a force A, acting 
singly, would cause a particle at .4 to describe the line AB 

in a time t, and a force F 2 , 
also acting singly, would 
cause the particle to pass over 
AD in the same time ; then, 
if they act jointly on the 
same particle, they will cause 
it to describe the line AC, 
PlG - 1S - the diagonal of a parallelo- 

gram, constructed on AB and AD, in the same time. 




f53-£6. ] KKSOLVED FORCES. 25 

This follows directly from the Second Law. For, the 
force F> will cause the particle to reach a line DO, which 
is parallel to A />, in the same time, whether Y'j acts or 
not; and similarly ./'J will cause it to reach a line DO, 
parallel to AD, in the same time ; and hence, when (hey 
act jointly, it will he found at the intersection of DO 
and BO, or at O, at the end of the time; and, as it must 
move in a straight line, it will move along the diago- 
nal AO. 

This is nearly the same as Article 14, only here we con- 
sider the cause of the motion, while there the velocities 
only were considered. 

53. Resolved Forces. — If a force F would cause a 
particle to describe the line AO uniformly in a given 
lime, it may he resolved into two forces, 1<\ and F z , one of 
which will cause it in the same time to describe the side 
A />\ and the other, AD of a parallelogram, of which AO 
is the diagonal. This is the converse of the preceding 
article. The result, however, is indeterminate, since AO 
may be the diagonal of an indefinite number of parallelo- 
grams. 

54. Components. — The forces jp[ and F 2 are called 
component forces ; and F is the resultant. 

55. The Resultant of two or more forces which act 
upon a single particle, is a force which will produce the 
same effect as all the other forces combined. 

56. Relation between two Forces and their Re* 
sultant. 

Let 

F= AO; F x = AD = DO; F Z = AD; 

a = OAD ; B = GAB ^AOD; 

= DAB = l$0° -ADC. 
2 



26 



KINETICS. 



[57,68.] 



The triangle ADC gives 

AD DC 



AC 



or, 



sin ACD~ sin DA C " sin ADC 9 

sin /3 sin a sin 6 ' 

57. Rectangular Components. — Let the components 
F[ and F 2 make a right angle with one another; and 

a and /3 be the same as in the 
preceding article. Then the 
triangle AD C gives 

F X = F cos /3; 

F 2 = F cos a = F sin /3. 

Squaring and adding, observing 
that sin 2 a + cos 2 a = 1, we have 

F* = F? + F 2 \ 




Fig. 14. 



58. If several forces act along the same line upon a 
body, whether in the same or opposite directions, theuc 
resultant is the algebraic sum of the several forces. 

Let R = the resultant, then 

B = XF 

58a. Concurrent Forces are those which act upon a 
body at the same point. When applied at different points 
they are non-concurrent. 

EXERCISES. 

L When a body is thrown horizontally into the air, why does it fall to- 
wards the earth ? (See the Second Law.) 

2. The planets are free bodies in space ; how much force does it require 

to deflect them from their course ? 

3. A stone, whose weight is 500 pounds, rests upon another stone whose 

weight is 2,000 pounds ; what is the reaction of the latter against 
the former ? 



[59,60.] CONSTANT FORCE. 27 

4. If two boats of equal size, offering equal resistances to movement on 

water, are connected by a rope, and a man in one of the boats pulls 
on the rope, drawing the boats toward one another, at what point 
between them will they meet ? If a man in the other boat also 
draws in the rope, at what point will they meet ? 

5. If a person is on ice, which moves in a due easterly direction at the 

rate of 1 mile per hour, and he walks on it at the rate of 3 miles 
per hour, in what direction must he travel so that his resultant 
course shall be due south ? 

6. If a ship is sailing south-east at the rate of 8 miles per hour, and the 

tide is carrying it due east at the rate of 3 miles per hour, what is 
its actual course, and its velocity ? 

7. A ship is sailing due south-east at the rate of 10 miles per hour ; what 

is its velocity in a southerly direction ? 

8. If a ball is placed on the floor of a railroad car, and is perfectly free 

to roll, will it change its position if the velocity of the train is 
gradually increased ? "Why would it rush toward the forward end 
in case of a collision with another train ? 

9. The force F\ equals 20 pounds, F 2 equals 20 pounds, and the angle 

between them is 60 degrees ; required the value of their resultant 
and the angle between it and the forces F x and F a respectively. 



Constant Force. 

59. A constant force is one which acts with a constant 
intensity. 

An incessant force is one which acts constantly, but with 
a variable intensity. 

60. A constant force applied to a free body, and acting 
along the line of motion, produces a constant acceleration. 

For, according to the first law, if no force be applied 
the velocity will remain constant, and, according to the 
second law, the change of velocity will be proportional to 
the force ; and as the force remains constant the change 
of velocity must be the same for each unit of time, that is, 
constant. 

If the force be applied so as to produce an increase of 



28 KINETICS. [61,62.] 

velocity, the acceleration will be positive ; but if it causes 
a decrease of velocity, the acceleration will be negative. 
In the former case the velocity would increase indefinitely, 
but in the latter case the body might be brought to a state 
of rest. For instance, a body thrown horizontally on ice, 
or a train of cars, moving after the locomotive is detached, 
would be brought to rest by friction. An active force, 
operating to destroy the velocity, may, by its continued 
action, produce a velocity in an opposite direction. Thus, 
when a train of cars is in motion, the locomotive may be 
reversed and push against the train to stop it, and by con- 
tinuing its action may finally produce a velocity in an 
opposite direction. 

61. A constant moving force is one in which the 
resultant of all the forces is constant. Thus, when a loco- 
motive draws a train of cars, it may exert a constant mov- 
ing force for a time ; that is, there will be a constant force 
for producing motion ; but, as the velocity increases, the 
resistance of the air increases, and usually, after a short 
time, the whole power of the locomotive is exerted in 
overcoming the friction and resistance of the air, and pro- 
duces no increase of velocity. In the latter case, although 
the locomotive may exert a constant force, it is not called 
a constant moving force, for the pulling force of the loco- 
motive is exactly neutralized by the resistances of the 
train. 

62. The movement of a body under the action of a 
constant moving force is illustrated by Fig. 9, because 
the force produces a constant acceleration. Hence, the 
formulas of Article 24 give the relations between the 
time, space and velocity, when the acceleration is known. 
The line of action of the force is supposed to be in the 
direction of motion. 



[63-65.J CONSTANT FORCE. 29 

63. Normal action. — If a force acts constantly normal 
to the path described by the body, it will not affect the 
velocity. Thus, if a body be connected to a point by a 
string, so that the body will describe the circumference of 
a circle, the tension of the string will not change the 
velocity. The force of the string and the velocity of the 
body may both remain constant. 

64. The force of gravity is one of the forces of nature. 
It is an attractive force, tending to draw bodies towards 
each other. It always manifests itself wherever there is 
matter. It is the force which gives weight to bodies, and 
causes unsupported bodies to fall to the earth. It holds 
the planets in their orbits. It acts through bodies with- 
out being diminished in its intensity, and upon the most 
central portions of a body with the same intensity as if 
the external portions were removed. 

65. The Law of Universal Gravitation is as follows : 
Two particles attract each other with a force which varies 
directly as the product of their masses and inversely as 
the square of the distance between them. 

This law was discovered by Sir Isaac Newton in 1666, 
but, on account of an erroneous value of the diameter of 
the earth which was then used, he was not able to prove 
it at that time. But in 1682 it was found, from new 
measurements of an arc of the meridian, that the correct 
diameter was about ^ greater than the value which he 
had previously used;. and with this value he fully de- 
monstrated the law. 

Let M = the mass of a body A ; 
m = the mass of a body B ; 
D = the distance between them ; 



30 KINETICS. [66, 77. J 

then the attractive force between will vary as 

mM 

66. The force of gravity at any place on the earth 
remains constant. — This is shown by the fact that the 
weight of a body is alway§ the same at the same place ; 
also, that a body always falls the same distance in a 
vacuum in the same time at the same place. 

It is well known, however, that upheavals are taking 
place in some parts of the earth and depressions in others, 
and these doubtless produce exceedingly slight changes in 
the weight of a body, but no apparatus at the present day 
is sufficiently delicate to measure these changes, if they 
actually exist ; hence, we may say that the force of gravity 
is at least practically constant at every place. 

67. Determination of the acceleration produced by 
gravity on a body falling freely in a vacuum. — We 
6peak of a vacuum because it is shown experimentally 
that ail bodies fall the same distance in the same time in 
a vacuum ; and also because it is necessary to exclude the 
resistance of the air in determining the full effect of 
gravity. This is a problem the exact solution of which 
involves considerable knowledge of mechanical principles 
and great skill in making observations. These principles, 
so far as they involve the use of the pendulum, will be 
explained hereafter, but in this place we can only describe 
the process. 

By means of Atwood's machine an approximate value 
of the acceleration may be determined. 

By means of delicate machinery, and a refined system 
of making observations, the space and time may be 
observed directly. 



[6a] CONSTANT FORCE. 31 

But the most reliable method, or, at least, that most 
commonly used, is by means of a pendulum. Any body 
vibrating on an axis, under the action of the force of gravity, 
is called & pendulum. If the vibrating body has percepti- 
ble size, it is called a compound pendulum. A simple 
^pendulum is a material particle suspended on a line 
without weight, and hence, is an ideal pendulum, but it 
has a real mathematical signification. A simple pendu- 
lum may always be found which will vibrate in the same 
time as a compound one. 

If I = the length of a simple pendulum ; 
t = the time of one vibration in seconds ; 
g = the acceleration due to gravity in a vacuum ; 
7T = 3.141592 = the ratio of the diameter of a circle 
to its circumference ; 
then 



— Vf» 



from which we find 

To determine t, the number of oscillations for a given 
time, say 10 or 20 minutes, is observed, and this number, 
divided into the number of seconds in the time, gives t. 
The length I is measured directly. These quantities, sub- 
stituted in the preceding formula, will give the value of g y 
which in this latitude is about 32J- feet. In this way the 
intensity of gravity has been found at different places on 
the earth's surface. 

68. The intensity of gravity varies with the lati- 
tude. — By means of the experiments indicated in the 
preceding article, it has been found that the force of 



32 KINETICS. [69,70.] 

gravity is least at the equator, and increases with the lati- 
tude both north and south of the equator. The value of g 
for any latitude L may be found approximately by the 
formula 

g = 32.1726 - 0.08238 cos 2Z; 

but no formula will give the exact relation between g and Z. 
At the equator L = 0, and at the poles L = 90° ; hence we 
have 

at the equator g = 32.0902 feet, and 

at thejpoles g % = 32.2549 feet. 

69. Weight variable. — According to the preceding 
article, a body will weigh less on the equator than at any 

other place on the surface of the earth. But 
/**\ this difference could not be detected by a com- 
mon beam balance, for, a diminution of the 
weight at one end of the beam would be exactly 
the same as that at the other, and if the bodies 
balanced at one place on the surface of the 
earth they would balance at every other place. 
The difference, however, might be detected by 
a spring balance. Fig. 15, for the more the body 
weighed the more it would pull the index down. 
fig. is. If a body be elevated one mile above the sur- 

face of the earth, it will, according to Article 
65, lose about ^ of its weight. These variations being 
small, we may, for the purposes of this work, consider g as 
constant, and equal to 32-J- feet. 

70. There are three causes for the variation of 
gravity on the surface of the earth. — First, the earth is 
an oblate spheroid, the axis of which coincides with the 
axis of the earth ; and those bodies on the equator, being 
more remote from the centre of the earth than those at 



[71.] 



CONSTANT FORCE. 



33 




higher latitudes * will be attracted with less intensity ; 
second* the revolution of the earth on its axis produces a 
so-called " centrifugal force " (to 
be explained hereafter), which 
tends to throw bodies from the 
surface, thus diminishing their 
weight ; and third, the heteroge- 
neous character of the substance 
of the earth. 

The form of the earth is sup- 
posed to be due to the attraction 
of the particles for each other 

and to the centrifugal force caused by the rotation on its 
axis while the substance of the earth was in a plastic state. 

71. The atmosphere resists the movement of bodies in 
it ; and hence, the velocity of bodies under the action of 
any force is less than it would be in a vacuum. The 
attraction of the earth being the same on each particle of 
a body, a light body would fall as rapidly as a heavy one 
if there were no resistances to their movements ; and this 
is confirmed by experiment, by letting bodies fall in a 
vacuum. The resistance of the air varies with the surface 
against which it acts, but in falling bodies the ability to 
overcome this resistance varies as the weight of the body ; 
hence, heavy bodies fall faster than light ones in the air. 
"But the velocities of heavy bodies, such as iron, stone, 
brass, etc., falling 100 to 200 feet, do not differ much from 



• As determined by The semi-polar axis is and the equatorial radius in 

Bessel 20,853,662 feet ; 20,923,596 feet. 

Aiiy . . , 20,853,810 feet ; 20,923,713 feet. 

Clark© 20,853,429 feet ; 20,923,161 feet. 

The equatorial diameter of the earth is about 26 miles longer than its 
axis. 

2* 



34 KINETICS. [72-75.] 

each other; and for compact masses of such materials 
falling in air we use 32£ feet for g. 

Formulas for Falling Bodies. 

72. Bodies falling from rest. — The acceleration being 
constant, we have only to substitute its value in the equa- 
tions of Article 24. Making f=g and s = h in those 
formulas, we have, 

v = gt; . . . . . (1) 
h = \gfi; . . . (2) 

.'.* = *«* = ^; . . . (3) 

, , v /2A\* 2A 
and t = - = \— I = — . . . (4) 
g \g/ v 

73. If a body is projected downward with a velocity 
v , we makef = g and s = h in Article 25, and have 

v = gt + v Q . . . . (5) 

h = ig# + v t. . . . (6) 

74. If a body is projected upward with a velocity, 
the acceleration becomes negative, and the equations of 
the preceding article become 

v = v -gt. . .. . (7) 

h = v t-igt?. . . . (8) 

75. Problems. — 1. If a body is projected upward with 
a velocity of 100 feet per second, required its height at 
the end of 2 seconds. 

Equation (8) gives 

h = 100 x 2 -i x 32£ x 4 = 135f feet. 
2. In the preceding problem what will be the height at 
the end of 8 seconds ? 
We have 



T75.1 FORMULAS FOR FALLING BODIES. 35 

h = 8 x 100 - i x 32| x 8* = - 229^ feet ; 
that is, at the end of 8 seconds, the body will be 229-J feet 
below the starting point. 

3. In Problem 1, what will be the greatest height of 
ascent ? 

When it is at the greatest height v will be zero in equa- 
tion (7) ; hence 

v = gt. 

.'. t = -° = 3.1 + sec. 

g 

and this in equation (8) gives 

h = 100 x 3.1 - i x 32£ x (3.1) 2 = 156.4 feet. 

±. If a body is projected upviard with a velocity of 200 
feet per second, required its height when the velocity ia 
100 feet per second. 

From equation (7), we find for the time 

200-100 600 
t = — - — = m seconds, 

which substituted in equation (8) gives 

EXAMPLES. 

1. A body falls from rest through a height of 100 feet; 

required its final velocity. (Let g = 32J feet.) 

2. A body falls from rest and acquires a velocity of 300 

feet ; required the time. 

3. A body is projected upward with a velocity of 200 feet 

per second* what will be the greatest height of 
ascent? 



36 KINETICS. [76.] 

4r. If g = 32£ feet per second, what will be the accelera 

tion per minute. 
5. A metre is 3.28 feet (nearly) ; if the unit of time were 

2 seconds what would be the acceleration % 
6 If a body is projected downward with a velocity of 25 

feet per second, what will be the velocity after it has 

fallen 120 feet 1 

7. In the preceding example, what will be the time of 

descending 150 feet \ 

8. At the instant a body is dropped from a point A, 

another body is projected upward from a point B, 
vertically under A, and they meet at the middle of 
AB\ required the velocity of projection from B. 

Ans. (AB x g) x \ 

9. A body is let fall into a well, and 4 seconds afterward 

it is heard to strike the water ; if the velocity of 
sound is 1130 feet per second, required the depth of 

the well. 

Ans. 231 feet. 

10. A body is projected downward from a point A with a 

velocity of v feet per second, and another body is 
projected upward from a point a feet below the 
former, with a velocity of V feet per second; 
required their point of meeting. 



EXERCISES. 

1. If a boy draws a load on a sled with an increasing velocity, will he 

exert any more force than if he draws it at a uniform rate ? 

2. Why will a body fall more rapidly at the foot of a mountain than at 

its top ? 

3. Why may some light bodies fall more rapidly at the top of a moun- 

tain than at its foot ? 

4. If a body, whose weight is 5 lbs. , attracts a particle at a distance of 

5 feet with a force of tott of an ounce, with what force will 



176.J ATTRACTION OF SHELLS. 37 

another body, whose weight is 25 lbs. , attract the same particle at 
a distance of 15 feet ? 

5. Which will vibrate in a shorter time, a pendulum 10 inches long or 

one 15 inches long? 

6. If a pendulum vibrates once each second at New York, will the time 

of a vibration of the same pendulum be more or less than a second 
at the equator ? 

7. If a merchant weighs iron in New York and sends it to some port 

near the equator ; will it gain cr lose in weight if it be weighed in 
both places with the same beam scales ? Will it gain or lose if 
weighed with the same spring balances ? 

8. What is the value of g (32£ feet) in metres per second ? 

Attraction of Homogeneous Shelh. 

76. Problem. — The attraction of a perfectly homoge- 
neous, spherical shell is the same upon a particle placed 
anywhere within it. 

Let ABODE he a section of an indefinitely thin spheri- 
cal shell, and any point within it. Draw lines BOD 
and AOC through the point 0, mak- 
ing an indefinitely small angle with 
each other, and consider A OB and 
00 D as two cones having their verti- 
ces at O, and their bases AB and 
CD in the surface of the sphere. The 
quantities of matter in each will be 
directly as their bases, but at the limit 
the triangles are similar, and the bases will be as A O 2 to 
OC 2 . The attraction of each varies as the quantity of 
matter and inversely as the square of the distance from ; 
(see Article 65) that is, 

Attraction of AB ^ A0 ^ * (AOJ 2 = 1 
Attraction of CD ' „ „ m 1 5 




KINETICS. 



f 77, 78. 1 



hence, the attraction of the bases will equal one another, 
and, being in opposite directions, will neutralize one 
another's effect. Similarly, conceive that the shell con- 
stitutes the bases of an indefinitely large number of cones, 
then, according to the above reasoning, the attraction of 
all the matter on one side of any straight line drawn 
through will exactly neutralize that on the other side. 
The same may be proved for any other point. 

77. Problem. — If the earth were a homogeneous, hollow 
sphere of uniform thickness, a body placed at any poi?it 
within the hollow would remain at rest. 

For, according to the preceding article, the attraction of 
any of the spherical shells of infinitesimal thickness upon 
any point within it will be zero, and 
hence, the effect of all of them upon 
the same point will be neutralized. 

It follows from this that if the earth 
were solid and composed of homogene- 
ous, concentric shells, though they var- 
ied according to any law from the centre 
to the surface, the resultant attraction 
upon a particle at the centre of the earth would be zero. 

78. Problem. — If the earth were a homogeneous solid 
sphere, the resultant attraction upon any point within it 
would vary directly as its distance from the centre of the 
earth. 

Suppose that a particle is at a distance x from the cen- 
tre ; then, according to the preceding problem, the result- 
ant attraction of the shell outside of x will be zero ; and 
that portion of the sphere whose radius is x will attract 
directly as its quantity of matter and inversely as the 
square of the distance of the particle from the centre of 
the sphere. 




Fig. 18. 



[79-81.] WEIGHT — MASS — DENSITY. 39 

The quantity of matter will be directly as the volume, 
or as 

and hence, the attraction will vary as 

■| 7r as 3 -f- x 2 , or as -§- ir x ; 

that is, directly as the distance of the particle from the 
centre. 

Weight — Mass — Density. 

79. Weight. — The weight of a body has already been 
defined, in Article 32, as a measure of the attractive force 
of the earth upon the body. 

Weight is not essential to matter. According to Arti- 
cle 76, if a body were placed at the centre of the earth it 
would weigh nothing. Similarly, if placed at a certain 
point between the earth and moon it would lose its weight. 
According to the preceding articles its weight depends upon 
its position ; or more specifically, upon the attraction of the 
earth upon it. 

80. Mass is a term used to express quantity of matter. 
—This, as we have seen, is independent of the weight of 
the body ; in other words, it is constant for the same body. 
The ratio of the weights of two bodies in vacuo, deter- 
mined at the same time, whether on the equator, the top 
of a mountain, within the earth, or at any place in the 
universe, is constant.. Hence, if two bodies, one weighing 
2 pounds and the other 10 pounds, be balanced on a lever 
at any place, they would remain balanced if taken to a 
place where the bodies weighed 1 and 5 pounds respect- 
ively, or 4 and 20 pounds respectively. 

81. Measure of Mass. — The mass of a body equals its 



40 KINETICS. [81.J 

weight at any place divided by the acceleration due to 
gravity at the same place. 

1„ must be measured in such a way as to give the same 
value wherever determined. As already shown, the weight 
varies directly as the force of gravity, and the acceleration 
due to gravity also varies as the same force ; hence, the 
ratio of the weight to the acceleration will be constant fcr 
all places, both being determined at the same place. 
If 

M = the mass of a body ; 
W = the weight of the body at any place, and 
g = the acceleration due to the force of gravity 
at the same place ; 
then, 

if oc , 

9 

which may be put in the form of an equation by intro- 
ducing a constant c, hence, 

M=c — . 
9 

Since c is arbitrary, the unit of mass may be so chosen 
that c will be unity, in which case we have 

ir w 

which is the value used in Mechanics. 

The weight in this case must be determined by a spring 
balance, or its equivalent, which must weigh correctly a 
standard pound, or multiples thereof. 

For ordinary practical purposes it is only necessary to 
divide the weight of a body, determined at any place on 
the earth with a pair of good scales, by 32£. 



[82-83.] WEIGHT — MASS — DENSITY. 41 

The mass of a body might also be found by weighing it 
at any place with a pair of beam scales, using as weights 
standard units of mass ; that is, 32-j- standard pounds of 
weight. 

82. The mass of a body is the number of pounds that a 
body would weigh at that place where the acceleration due 
to gravity is one foot per second, the weight being deter- 
mined by a standard spring balance. 

In the last equation of Article 81 make g = 1, and let- 
ting Wi be the corresponding weight, we have 

M=W 1 . 

One such place is in the earth, at about -^ of the radius 
of the earth from the centre, or less than 125 miles from 
the centre of the earth. Another point is about 22,000 
miles from the centre ; for if x be the distance, and the 
radius of the earth be called 4,000 miles, we have, accord- 
ing to Article 65, 

(4000) 2 : x 2 : : 1 : 32£ ; 
.-, x = 22,686 miles. 
Since the weight of the same body varies directly as the 
force of gravity, we have 

M pounds : W "pounds : : 1 foot : g feet ; 

9 
In this expression g may be considered as an abstract 
number, being the ratio of the acceleration due to gravity 
at two different places. 

83. The Unit of Mass is a body which weighs Z%\ 
standard pounds. 

For, if M = 1 in the preceding article, we have 
W pounds = g x 1 pound = 32^ pounds. 



42 KINETICS. [84,85.] 

The term pounds is used in a double sease. We use 
pounds of weight and pounds of mass, but no ambiguity 
arises on account of it, for the language of the problem 
will always determine which lis referred to. 

Since the weights of bodies at the same place are 
proportional to their masses, the mass of a body may be 
found by comparing its weight with that of a body whose 
mass is definitely known. This is done by means of a 
simple balance. 

84. Analytical Expression for Weight. — From 
the last equation of Art. 82 we have 17= Mg\ an ex- 
pression which is useful in the solution of many problems. 

85. Density relates to the compactness of matter. If 
the mass of a body be uniform throughout the volume, 
the density is the mass of a unit of vohime. 

Let M= the mass of a homogeneous body, 7 = the 
volume of the body, D = its density ; then 

M 
& = y -:M=DV. 

If 7=1, then 

Z> = M. 

Substitute the value of M from the last equation of 
Article 82, and we have 

W 

gY 

If the density he variable the density at any point of a 
body will equal the mass of a unit of volume having the 
same density throughout the unit as that at the point of 
the body considered. 

The unit of volume may be taken as a cubic inch, foot, 
or any other standard measure. 



(.85 J MEASUEE OF FORCE. 43 



EXEECISES. 

1. If the earth were a homogeneous sphere, and a body on its surface 

weighed 10 lbs., what would be its weight if placed at a point 
half way 'between the surface and centre ? 

2. If a body weighs 10 lbs. on the surface of the earth, what would it 

weigh if elevated to a point above the surface equal to the radius 
of the earth ? 

3. If the earth were a homogeneous spherical shell of finite thickness, 

what would be the weight of a body placed anywhere within the 
hollow ? 

(The earth is referred to in these questions because the condi- 
tions could not be realized by a hollow sphere on the surface ; for 
gravity would exert its full force on a body placed in the hollow 
of such a sphere. ) 

4. In the preceding example, if a body weighed 10 lbs. on the surface 

of the earth, at what place in the shell must it be placed that its 
weight shall be 5 lbs. ? 

5. If the earth were a homogeneous shell, and a body were dropped 

from the surface into the hollow, would the motion be accelerated 
or not as it passes through the shell ? And as it passed across 
the hollow would its motion be accelerated or not, no allowance 
being made for the resistance of the air ? 

6. tf a person were placed in the hollow described in the preceding 

question, and should jump from one side toward the centre of 
the sphere, where would he stop ? Could he stop at the centre if 
he desired to ? 

7. If a person were placed at one extremity of a diameter of the hollow 

referred to in exercise 5, and a ball of equal mass placed at the 
other extremity, and the person should pull on the body by means 
of a string, where would they meet ? If he pulls for an instant 
and then ceases to pull, will they meet ? If he pulls with a constanb 
foice until they meet, will their acceleration be uniform or 
variable ? 

8. In the preceding question, if the ball has half the mass of the person, 

which will move faster, the ball or the person ? 

9. If a person were placed at the centre of the hollow sphere of exer- 

cise 5, and not able to reach anything, could he move away from 
the centre by his own exertions ? If he had a ball and should 
throw it away, would the person move away from the centre ? 



44 KINETICS. [88,87.1 

(He could not drop the ball, for it would not move in any lirection 
unless started. ) What would be their relative directions of motion ? 
Would they move in straight or curved lines ? 

10. If a body weighs 100 standard pounds, how many pounds of mass 

does it contain ? 

11. If a body whose volume is 2 cubic feet weighs 200 lbs., what is 

its density ? 

12. If a body weighs 5 kilogrammes, what is the mass in pounds ? 

13. If the weight of a body whose volume is 1 cubic metre is 3 kilo- 

grammes, what is the density in pounds per cubic foot ? 

14. If a hole were made through the centre of the earth from surface 

to surface, and a ball were dropped into it, would it stop at the 
centre ? Where would it stop if the hole were a vacuum ? At 
what point would it move with the greatest velocity ? 



Dynamic Measure of Force, 

86. Value of a Moving Force. — Forces are compared 
by their effects. If a force jp] acting as a constant pull 
or push on a body perfectly free to move in the direction 
of action of the force, produces an acceleration f and 
gravity acting on the same body with a constant force W } 
equal to the weight of the body, produces an acceleration 
g, we have 

F: W::f:g 

,:F= W^=Mf; 

that is, a constant moving force of F pounds equals the 
product of the mass into the acceleration in feet per 
second. 

From the last equation we have 

■f-H- Z 
* ~ M — W 9 ' 

87. Effective Force. — Only a portion of the forces 
which act upon a body, may be effective in moving it, the 



[88,80.] MEASURE OF FOBCE. 45 

others neutralizing each other. Thus, when a locomotive 
draws a train of cars, a portion of the pulling force is 
directly neutralized by the resistance of the air, friction 
on the track, and other resistances of the train. If the 
pulling force exceeds the resistances, the excess will he the 
effective pulling force. When the resistances equal the 
pulling force the motion will be uniform. 

88. Remark. — To find the space passed over by a body, 
and the velocity produced in a given time by the action of 
a constant effective force, find the value of f from Article 
86, and substitute its value in the equations of Article 24. 

89. Problems. — 1. If a piston he driven apportion of 
the length of a cylinder hy a constant steam pressure, at 
what point must the pressure he 

instantly reversed so that the full 
stroke shall just equal the length 
of the cylinder, the cylinder heing 
horizontal, and the piston moving 
without frictio n f 

At the middle of the stroke. 
Whatever velocity is generated by the action through one 
half of the stroke will become neutralized by the same 
pressure acting in the opposite direction through the 
remaining half. 

2. In the preceding example, what will he the velocity 
at the centre of the cylinder ? 

Let F = the total pressure of the steam, 
W = the weight of the piston, 
s = one-half the length of the cylinder ; 
then, from Article 86, we have 

F 

f= w e; 





46 KINETICS. [89.) 

anu, if the piston starts at one end, we substitute this 
value in equation (3) of Article 24, and find 

3. A body whose weight is W is moved horizontally on 
a frictionless surface by the pull of a constant force F; 

required the space passed over 
JBLM f- in a time t, and the velocity 

acquired. 

Here we have, 
F 

hence, from Article 24, we find 
and S -2TF*' 

LIfa body, whose* weight is W on the surface of the 
earth, be placed in the hollow sphere described in Article 
76 ; required the constant pull or push necessary to pro- 
duce a velocity v in time t. 

From Article 86 we have 

* F 

which, substituted in equation (1) of Article 24, gives 

F \ 



from which we find 






L90-1 MEASURE OF FORCE. 47 

EXAMPLES. 

1. If a piston weighs 100 lbs., and the constant steam 
pressure is 2,000 lbs., what will be the velocity 
acquired in moving over 12 inches ? 
In the third problem above, if the weight of the body 
is 500 lbs., and the constant pulling force is 25 lbs., 
required the space over which the body will be 
moved from rest in 10 seconds. 

3. In the same problem, if the constant frictional resist- 

ance is 10 lbs., what will be the velocity at the end 
of 100 feet? 

4. In the fourth problem above, if a body weighs 100 lbs. 

at a place where g — 32 feet, and is placed in a 
hollow space at the centre of the earth ; required 
the pulling force necessary to produce a velocity of 
100 feet in 10 seconds. 

90. Problems. — 1. Suppose that a body is on a friction- 
less plane, and is moved horizontally by a weight attached 
to it by means of a string pass- 



vr 



ing over a pulley ; required 
the space passed over by each 
of the bodies in a time t, there 
being no resistances from the 
pulley, string or air. 

P = the effective moving force, 

P+ W 

= the total mass moved ; 

9 
hence, from Article 86, we have 

P P 

f = P + W :== P + W g; 



M 



48 KINETICS. [90.] 

and this, in equation (4) of Article 24, gives 

(Remark. — If the student finds it difficult to distinguish letween 
the moving force and the mass moved, let him imagine the whole 
system placed in the hollow sphere described in Article 76. Then both 
bodies, P and TT, will be destitute of weight, and no motion can result 
from the action of P. Now conceive that a string is attached to the 
body 2', and passed through the shell to some point on the surface 
where a man pulls with a constant force of P pounds ; the result will 
be the same as that given in the preceding problem, for the man will 
be obliged to move the mass of both bodies. When the system is on 
the earth, gravity pulls with a force of P pounds on the body P, and 
nothing in the direction of motion of W\ hence the force P must move 
both P and W.) 

2. Required the tension of the string m the preceding 
problem. 

Let T = the tension. 

Conceive the string to be severed, and a force applied 
equal to the tension T, pulling upward on the 
body ; then will the effective moving force be 
(P- T) pounds. 
The mass of the body P, will be 

Pio. 82. _ . 

hence, according to Article 86, we have 

p 

g J 

The effective pulling force on W, Fig. 21, is the tension 
of the strings hence, according to Article 86, 

9 



[90.] MEASURE OF FORCE. 

Eliminating f from these equations gives 

WP 



49 



T = 



W+ P 



3. A string passes over a wheel and has a weight P 
attached at one end and W at the other • if there are no 
resistances from the string, wheel, or air, 
and the string is devoid of weight, required 
the resulting motion. 

The effective moving force F, — (W— P) 
pounds ; 

the mass moved = ; 

9 



hence, from Article 86, 



W 4- P 




Fig. 23. 



-•_w-p 

"J - W+P g; 
and, from Article 24, we have for the space 
, W-P 



8 = 

and for the velocity, 



W+P 

W-P 
W+P 



gf, 



gt. 



(These problems are ideal, since they discard certain elements, such 
as the mass of the pulley, friction, and stiffness of the cord. These, 
however, may all be computed, but they make the problem too compli- 
cated for this part of the work. The chief object here is to confine the 
attention to the relation between forces and the motion produced upon 



50 KINETICS. [90.J 

4. Find the tension of the string in the preceding 
problem. 

2PW 



Ans. T = 



P+W 



5. A. inan, whose weight is W, stands on the platform 
of an elevator as it descends the vertical shaft of a mine ; 
if the platform descends with a uniform acceleration of 
\g, show that his pressure upon the platform is £ W. 
What will it he if the platform ascends with the same 
uniform acceleration f 

EXAMPLES. 

1. In Fig. 21, if P = W, how far will the bodies move in 

5 seconds % 

2. In Fig. 23, if W = 2P, how far will the bodies move 

in 5 seconds ? 

3. In Fig. 23. W = 50 pounds ; what must be the 

weight of P so that it will descend 10 ft. in 5 
seconds? What, that it may ascend 10 ft. in 5 
seconds ? 

4. In Fig. 23, if P — 5 ounces, and W = 4 J ounces, 

and it is observed that P descends 6.8 ft. in 2 
seconds ; required the value of g. 

5. In Fig. 23, if W= 10 pounds, required the weight of 

P that the tension may be 1 pound ; 5 pounds ; 10 
pounds ; 20 pounds. 

6. In Fig. 21, if W = 20 pounds, P = 2 pounds ; re- 

quired the time necessary for the bodies to move 
10 ft. 



[90.] MEASURE OF FORCE. 51 



EXERCISES. 

1. A spring balance may be inserted in the string of the preceding 

problems in such a way as to indicate the tension. Suppose that 
such a balance were inserted in the vertical part of the string in 
Fig. 21, and the string cut off above it, what will the balance indi- 
cate afterwards ? 

2. If a man pulls a weight vertically upward by means of a cord 

attached to the body, will the tension on the cord equal or exceed 
the weight of the body ; — the weight of the cord being neglected. 
Consider the case when the motion is accelerated, and when it is 
uniform. 

3. If a man stands on the platform of an elevator as it descends a shaft 

with an acceleration, how will his pressure upon the floor compare 
with his weight ? How if the platform is ascending ? 

4. If a man slides down a vertical rope, checking his velocity by grasp- 

ing it more or less firmly, but with a constant grip, how will the 
tension on the rope compare with his weight ? 



CHAPTER m. 

WORK FRICTION. 

91. Work is the overcoming of resistance continually 
recurring along the path of motion. — Thus a horse, while 
drawing a load on a cart, does work by constantly over- 
coming the friction of the axle and the resistances of the 
roadway. The same effort, however, may be exerted in 
producing motion only, producing live or stored work. 
Hence, the following is a more general definition : Work 
is the effect produced by a force in moving its own point 
of application in such a way that it has a component 
motion in the direction of action of the force, 

92. Measure of Work. — A horse that draws a load 
two miles does twice the work of drawing it one mile, and 
one-fifth the work of drawing it ten miles. The work, 
therefore, varies directly as the space over which the 
resistance is overcome. It is also evident that, if the horse 
had drawn a load twice as large, he would have done twice 
the work in the same distance ; hence, the work also varies 
directly as the resistance. This principle is general, and 
applies to all cases in which the force is constant. 

Hence, if 

F = a constant force ; 
s = the space over which F acts, and 
£T= the work done by F\ 
then 

U = Fs. 

93. The -work of a variable force is found by divid- 
ing the space into small parts, so small that the force 



[94-06.] WORK, 53 

may be considered constant over each part, and taking 
the sum of all the elementary works. 
Let 

F = the force acting over any one of the 

elementary spaces ; 
As = an elementary space ; 
then 

U= 2 Fa's. 

94. Mere motion is not work. — If the planets move in 
space without meeting any resistance, they do no work. 

95. The Unit of Work is the raising of one pound of 
matter vertically one foot, and is called a foot-pound. 
The resistance overcome by raising matter is the force of 
gravity. A weight of one pound drawn horizontally ia 
not the unit, unless the frictional resistance should happen 
to equal the weight. It is not the weight moved, but the 
resistance overcome, that constitutes work. 

96. Work represented by a Diagram.— When the 
force is constant the work may be represented by a 
rectangle. 

* P w 



D 

F 



Fxo. 34. Fig. 25. 

Thus, in Fig. 24, let the base of the rectangle represent 
the space s, and the altitude represent the force F, then 
will the area of the rectangle be 

Fs, 
which is the expression given in Article 92 for the work. 

If the force does not follow a known law, we may still 
find the work approximately by constructing a curve, the 
abscissas of which, Aa, ab, be, etc., shall represent the 




54 KINETICS. [97,99.] 

spaces, and the corresponding ordinates, am, In, co, etc., 
shall represent the corresponding resistances ; then the area 
of the figure will represent the work, for the area will be 
XFAs. The area may be found to any degree of approxi- 
mation by dividing it into an indefinite number of trape- 
zoids, finding the area of each and taking their sum. 
If the force varies directly as the space over which it 
acts, the work may be represented 
by the area of a triangle, of which 
the base represents the space and 
the altitude, the final force. For, 
any ordinate be will be directly as 
its distance from A. 

97. The total work is independ- 
ent of the time required to perform it. — For, the space 
may be passed over in a longer or shorter time without 
affecting the product of F into s. The horse that draws 
a load one mile, does a definite amount of work, whether 
it be done in one hour or five hours. 

98. Time and velocity are involved in doing work ; 
for, the space involves both. If the space be passed over 
with a constant velocity we have 

* = vt, 
in which 

v = the velocity, and 

t = the time. 

Hence, U=Fs = Fvt; 

therefore, if the velocity be uniform, the work will vary 
directly as the time ; and if the time be constant, the work 
will vary directly as the velocity. 

99. Dynamic Effect. — If the velocity be uniform and 
< = 1, we have work = Fv . 



[100-102.] WORK. 55 

This is the work done in a unit of time, and is called 
Mechanical Power, or Dynamic Effect, and sometimes 
simply Power. It is the rate of doing work, or simply 
work-rate. 

100. The unit of Dynamic Effect is called the horse- 
power. It is 33,000 pounds raised one foot per minute. 
It was determined by Boulton & Watt by means of trials 
with horses at a colliery in England. It is doubtless much 
larger than what the average of good horses can do for 
hours in succession, but it may be considered as an arbi- 
trary unit, by which the work done by a horse or othei 
working power in a given time, may be measured. 

101. Work may be useful or prejudicial. — That is 
useful which produces the article sought, and that which 
wears out the machinery is prejudicial. The former 
produces money for the mechanic while the latter costs 
money. 

Prejudicial work is generally frictional in its character, 
but all friction is not prejudicial. Thus, the friction be- 
tween the driving wheels of a locomotive and the track is 
necessary, and hence, useful. In a similar way the fric 
tion between belts and pulleys is useful. 

It is not always possible to 
distinguish between useful and ij 

prejudicial work. The latter -^^ 

always accompanies the former, |\_ 

but we know that, for economy, jl toss ^ss^^^^^^^S J n 
the latter should be reduced as ^ ^ 

much as possible. 

102. If the force acts at an angle with the line of 
motion it may be resolved into two forces, one of which 
is parallel to the path described by the body and the other 



5Q KINETICS. [103,104.] 

perpendicular to it. Thus, in Fig. 27, the normal com* 
ponent of the force F is 

JV= F sm FOP; 
and the horizontal component is 

P = Fcos FOP. 
There being no motion normal to the path, the former 
component does no work ; and the work done by the latter, 
according to the definition, will be 

U= Ps = Fs cos FOP. 
If a body is moved around a semi- circumference by a 
constant force acting parallel to a fixed diameter, the 
work will be the product of the force into the diameter. 

103. Work in a Moving Body. — A moving body can- 
not be instantly brought to rest, and when both resistance 
and space are involved in a result, work has been done. 
The work which a moving body is capable of doing equals 
the product of the mean resistance which it overcomes 
into the space over which it works. Thus, if a cannon 
ball should penetrate the earth 10 feet, and the mean 
resistance were 500 pounds, the work done would be 5,000 
foot-pounds. Another mode of measuring it is given in 
Article 111. 

Friction. 

104. Friction is that force or resistance between two 
bodies which prevents, or tends to prevent, one body from 
being drawn upon another. 

All bodies are rough. However perfectly they may be 
polished, an inspection of their surfaces with a glass of 
high magnifying power, shows that they are still very 
rough. A smooth surface is a comparative term, implying 



[105.] FRICTION. 57 

that it is more or less smooth. A perfectly smooth surface 
probably does not exist ; but when the term is used it 
means that the surface offers no resistance of any kind. 
It is an ideal surface. 

It is certain that if two perfectly smooth plane surfaces 
were brought in contact, they would offer a great resist- 
ance to being drawn upon one another on account of the 
adhesion between the bodies. They would hold to one 
another nearly as strongly as if they were solid. When 
we refer to smooth surfaces in problems, this force is also 
excluded. 

Under every condition of one surface moving in con- 
tact with another surface no more friction is brought 
into play than is required to prevent motion. There- 
fore friction is a self -adjusting force. 

When one body is just at the point of sliding on an- 
other body the friction exerted is called limiting fric- 
tion. Its magnitude bears a constant ratio to the normal 
reaction ; it is independent of the shape and extent of 
the surfaces, and its direction of action (as motion begins) 
is opposite to the direction of motion. 

105. Experiments in regard to Friction. — M. Morin, 
a French savant, was one of the first to determine the laws 
of friction. These laws were deduced by experiments 
upon a variety of substances under a variety of conditions. 
A device similar to that shown in Fig. 29 was used. A 
body W was placed upon a long strip of another body, and 
it was desired to determine the friction between them. 
A strong fine cord was attached to the body, and, passing 
over a pulley at the end of the platform, was attached to 
a dish in which were placed weights P. The weights P 
were made to exceed slightly the frictional resistance, and 
thus pull the body W along the other body. The space 



58 



KINETICS. 



[106, 107.] 



over which they moved in a given time was then observed, 
and with the data thus obtained the friction was computed 
as shown in Article 109. 

Weights were added to the body W so as to produce 
greater pressure ; also weights were added to the dish and 
taken from it, so as to produce a greater or less velocity. 

Different substances were used for the body W, and 
also for the horizontal strip on which the body W slid. 

106. Angle of Friction. — Suppose that a body W is 
placed on an inclined plane A C, one end of which is 
gradually raised until motion begins, or the body is in a 
state bordering on motion. The weight TFmay be resolved 

into components ; one, parallel to 
the plane, which tends to pull the 
body down it, and the other per- 
pendicular to it. 

Draw Oa parallel to AC, and 
ah perpendicular to it; and let 
Oh represent W. Then, according 
to Article 52, we have 

F=Oa = TFsin ahO = TFsin CAB. 

This equals the resistance due to friction. 
Let 

iV= the normal pressure = ah ; 
then 

JST= TTcos ahO = Wcos CAB ; 
hence, 

F WsykCAB ^ „.- 

1\7 = w TTITB = tan CAB ; 

Jy TFcos CAB ' 

which is called the angle of friction, or angle of repose. 

107. Laws of Friction. — The following approxi- 
mate laws have been deduced from the experiments of 
Morin and others: 




Fig. 28. 



[108.] FRICTION. 59 

1. Friction of motion is slightly less than that of rest. 

2. The total amount of friction is independent of the 
area of the surfaces in contact. 

3. The amount of friction between two surfaces is 
dependent on the character of the surfaces in contact 
and varies directly as the normal pressure. 

4. Sliding friction is independent of the velocity. 

These laws are sufficiently accurate for ordinary ve- 
locities when the surfaces do not abrade, or cut one 
another. Lubricants are employed to diminish friction. 

To these may be added two exact laws, the first of the 
two being a direct result of the principle of equilibrium : 

5. If a body is in equilibrium, the friction is equal 
and opposite to the tangential component of the applied 
forces. 

6. If the tangential component of the resultant of the 
applied forces becomes greater than a certain limiting 
value, it cannot be balanced by friction. 

108. Coefficient of Friction. — According to the third 
principle of the preceding article, it follows that the ratio 
of the total friction to the total normal pressure between 
two surfaces is constant. This ratio is called the Coeffi- 
cient of Friction, 

Let JV = the normal pressure ; 

F= that force which is just sufficient to 
produce motion when acting par- 
allel to the plane of the surfaces ; 
it = the Coefficient of friction ; 
then 

F 



60 KINETICS. [109.] 

Comparing this with the equation of Article 106, we 
see that the coefficient of friction equals the tangent of 
the angle of repose. 

If the body moves on a horizontal plane, the normal 
pressure equals the weight , hence, 

F 

If W = 1, fi = F\ hence, the coefficient of friction 
equals the friction caused by one pound of the body. 

109. To find the value of the coefficient of friction 
of motion. 

Let W = the weight of the body on the plane, 
P = the weight which moves it, 
f= the acceleration, 
fi = the coefficient of friction ; 
then the total friction will be 

and the effective moving force, neglecting the motion of 

the pulley and the cordage, will be 

and the mass moved will 
be 
5} (W+P) + g; 

hence, according to Arti 
cle 86, 

r-^P+JTf. 



m 



Fig. 



P -fiW 
/ ~~ P + W 9 ' 



[109.] FRICTION. 61 

This value in equation (4), of Article 24, gives 

*~ a PTTT ff *' 

from which we find 

_ gfP- 2( P + W)s 

Knowing the weights P and W, and measuring care 
fully the space s and time t, the second member becomes 
completely known ; and hence, by reduction, /j, becomes 
known. 

Another method, which should be suggested by 
T 106, is to put a body on an inclined plane. Increase 
the inclination until the body is on the point of sliding. 
Since the coefficient of friction is equal to the tangent 
of the angle of repose, jj = tan # ? where 6 is the angle 
of inclination of the plane as the body is about to slide. 

Motion on a Rough Incline. — Referring to Fig. 28, 
let M be the coefficient of friction. It is required to 
find the acceleration of the body down the plane. The 
force F acting up the plane to retard motion is jaN. As 
shown in Art. 106, this force Fis equal to IF sin CAB, 
which for equilibrium 

ITsm CAB - M ^= 0. 

The force causing motion down the plane is the sum 
of the components along the plane A C. Hence, if f is 
the acceleration, 

IT sin CAB - M N= — t f- . . (1) 

if 

Resolving forces perpendicular to AC (see Art. 106), 
JY = TFcos CAB (2) 



61a KINETICS. [109a.] 

Eliminating N between (1) and (2), 

/ = g(sin CAB - ,ucos CAB). 



Show what the acceleration would be if the body were 
projected up the plane. 

109a. Experimental Determinations. — The range 
of results from experiments to determine the value of 
the coefficient of friction for different materials is indi- 
cated by the following values of M : 

Wood on wood, dry, .22 to .6 

". " " lubricated, .1 to .2 
Metals on metals, dry, .15 to .2 

" " " lubricated, .07 to .08 
For friction of rest, about 40 per cent may be added 
to the coefficients for friction of motion. 
Morin's results are given in the Appendix. 

M. Pambour made experiments upon the frictional 
resistance of trains of cars on some railroads in England, 
and found the friction to be 8 pounds per ton gross; 
hence the coefficient was 

8 i 

M ~~ 2240 ~~ W0 * 

Experiments in this country have given from 3£ to 6£ 
pounds per ton gross under favorable conditions and at 
low speeds. One of the more recent and accurate for- 
mulas is, resistance in pounds per ton =h J velocity in 
miles per hour -f- 2. Thus it may be noted that as the 
velocity increases the resistance increases. 

Lubricants are sometimes solid, as graphite and 
soapstone, but more often are liquid, of which there are 
many varieties. The lubricating oils are the more com- 
mon of this class, but water is sometimes used with 



[110.] FRICTION. 615 

economy. Tallow, soap, wax, etc., are classed as 
semi-solid lubricants. It is necessary to determine ex- 
perimentally the extent to which any particular lubricant 
will reduce the friction and wear for any special con- 
dition of service. The best and more common lubri- 
cants for usual conditions are, however, as follows : 

Light and delicate machinery : light mineral lubricat- 
ing oils, olive, clarified sperm, neat's-foot. 

Light pressures and high speed : sperm, refined 
petroleums, olive, cotton-seed, rape. 

Heavy pressures and high speed : heavy mineral oils, 
sperm-oil, castor-oil. 

Heavy pressures and slow speed : greases and solid 
lubricants. 

Yery heavy pressures : solid lubricants. 

Steam-cylinders : heavy mineral oils ; lard and tallow 
may be used. 

Shop machinery : heavy vegetable oils, heavy mineral 
oils, lard-oil, tallow-oil. 

[See Thurston's " Friction and Lost Work in Machinery and Mill- 
work. "] 

110. Problems. — 1. A piston is moved in a horizontal 
cylinder, as shown in Problem 1, Article 89, by a constant 
steam pressure of F pounds. At what point must the 
pressure be instantly reversed so that the full length of 
the stroke shall be a inches, there being a constant friction 
of F x pounds throughout the stroke? 

The effective driving force will be F — F x founds. 
The effective stopping force will be F + F x pounds. 
Let 

8 = the space over which the former acts, and 
8i = the space over which the latter acts ; 
then 

s + s t = a. 



62 KINETICS. [110.] 

The work done upon the piston by the driving force 
will equal that done by the stopping force ; hence, 

(F-F 1 )s = (F+F 1 )s l . 

Eliminating s 1 between these equations, gives 

F+'F X 
S = -2F^ a ' 

2. A stream of water falls over a dam hfeet high, and 
has a section of a square feet at the foot of the fall; 
required the horse-power constantly developed. 

(The section is taken at the foot of the fall, for the 
velocity is measured at that point.) 

The velocity at the foot of the fall will be 

v= V2gh, 

and the volume of water which passes over the dam in one 
seccnd will be 

a V2qh. 

The weight of a cubic foot of water being 62| pounds, 
the weight of the quantity will be 

W= 6%%a ^^gh pounds per second, 

= 60 x 62J& \/2gh pounds per minute. 

The weight, multiplied by the height h, through which 
it falls, will give the work it can do in one minute, and 
the result, divided by 33,000, will give the horse-power ; 

3. Find the worh necessary to d)*aw a body up an m 
dined plane. 



[110.] FRICTION. 63 

In Fig, 28, Article 106, let 
I = AC; b=AB; h = BC\ W= the weight of the body 
As shown in Article 106, the normal pressure will be 

JST= WcosCAB =jW, 
and hence, the friction will be 

which, multiplied by the length of the plane, gives the 
work necessary to overcome the friction ; 

jim = pub W. 
The component of the weight along the plane will be 

IF sin CA£ = jW, 

and the work of overcoming the weight will be I times 
this result, or 

hW; 

hence, the total work will be 

phW+hW; 

hence, it equals the work which would be necessary to 
draw the body horizontally from A to B, and lift it verti- 
cally from, B to C. 

4. Required the work necessary to compress a coiled 
spring a given amount. 

It is found by experiment that, as long as the elasticity 
of a spring remains perfect, the amount of compression 
varies directly as the compressing force. That is, if one 



64 KINETICS. [110.] 

pound compresses it one inch, two pounds will compress 
it two inches, and so on. Hence, if 

jp = the force which compresses a spring 

one inch, 
P = the total compressing force, 
s = the amount of the compression produced 

then 

P=jps, 

and the work is represented by Fig. 26, in which AB = s, 
BG=P. 

.-. U=%Ps = 



EXAMPLES. 

1. How many cubic feet of water will a 50 horse-power 

engine raise in an hour from a mine 500 feet deep, 
if a cubic foot of water weighs 62-J- pounds? 

2. Find the work necessary to raise the material in making 

a well 20 feet deep and 3 feet in diameter, if the 
material weighs 140 pounds per cubic foot. 

3. The pressure on a steam piston, moving horizontally, aa 

in Prob. 1, Art. 89, is 1,000 lbs., the friction 200 lbs. ; 
how far must the pressure act before it is instantly 
reversed that the full stroke may be 12 inches ? 

4. The French unit of work is one kilogramme raised 

vertically one metre ; required the equivalent in 
foot-pounds. (Take the metre at 39.37 inches and 
the kilogramme at 2.2 pounds.) 

5. According to Navier it requires 43,333 French units of 

work to saw a square metre of green oak; how 



[110.J EXERCISES. 65 

many foot-pounds will be required to saw a square 
foot of the same material ? 

6. A stream of water falls vertically over a dam 12 feet 
high, and has a transverse section of one square foot 
at the foot of the fall ; required the horse-power 
constantly developed. 

7 A hammer, whose weight is 2,000 pounds, falls verti- 
cally 8 feet; how far will it drive a pile into the 
earth if the constant resistance is 10,000 pounds ? 

8. If it is found by means of a spring balance that a span 

of horses pull with a constant force of 200 pounds 
in drawing a plough ; if they travel at the rate of 
2 miles per hour, what will be the mechanical power 
required to work the plough \ 

9. In Fig. 29, let W = 40 pounds, P = 8 pounds, and if 

is observed that P moves over 4 feet in 3 seconds ; 
required the coefficient of friction. 
10< In Fig. 29, W=2o pounds, P = 5 pounds, and the 
coefficient of friction = 0.15 ; required the space 
over which the bodies will pass in 5 seconds. 



EXERCISES. 

1. What is the unit of work ? Is work the same as force ? 

2. Is friction force ? When is it useful and when prejudicial ? 

3. In what sense is work independent of the time, and under what cir- 

cumstances is it dependent upon the time ? 

4. A body placed on a plane, which is elevated at an angle of 15 degrees, 

is just on the point of moving ; required the coefficient of friction 
between the body and the plane. 

5. A body, whose weight is 25 pounds, is on a horizontal plane ; re 

quired the tension of a string by which the body is drawn along 
uniformly, the coefficient of friction being £. 

6. Define mechanical power. Is it work ? 



CHAPTER IT. 

ENERGY. 

111. Energy is a term to express the ability of an agera 
to do work. We have seen, Article 103, that a moving 
body is capable of doing work. A slight consideration of 
bodies at rest shows that they are also capable of doing 
work. Thus, the water in a mill-pond is capable of doing 
useful work by being passed through a water-wheel. To 
do work the water must be in motion, but the weight of 
the water falling through a given height will do a certain 
amount of work, and this amount can be determined 
while the water is in position. Similarly, the same can 
be shown in regard to other bodies at rest. These ideas 
have given rise to the terms Kinetic energy and Potential 
energy. 

112. Kinetic Energy is the energy of a moving body, 
and is the work which the body must do in being brought 
to rest. It is visible energy. 

The work which a moving body is capable of doing is 
the same as if it had fallen in a vacuum through a height 
sufficient to produce the same velocity. 
Let 

W = the weight of a body, 
v = its velocity, 

h = the height through which the body must fall 
to produce the velocity v. 



[113.1 ENERGY. 67 

The work necessary to raise a body a height h will be, 
according to Article 92, 

Wh. 

If the body fall freely through the same height, it will 
be capable of doing the same amount of work when it 
reaches the foot of the fall. The velocity produced in 
falling a height h will be (Article 72), 

.= •^5 .-.*=£. 

Substitute this value in the preceding expression, and 
making, 

M-— (Article 80), 

we have, 

v 2 W 

Work = Wh = W^r = \ — tf = %Mv> = K. 
2 9 9 

The expression \Mv 2 is called the kinetic energy, and is 
represented by K, the initial letter of kinetic. It is also 
called the vis viva,* or living force of the body. Hence, 
the kinetic energy of a moving body equals the work 
stored in it. 

113. Potential Energy is latent energy. It is the work 
which a body is capable of doing in passing from one 
condition or position to another. Thus, the power in a 
coiled spring \& potential ; but, when freed from its restrain- 
ing power, it may move the wheels of a watch, or clock, or 
drive other machinery, in doing which it is changed from 
a condition of tension towards one free from tension. The 

* Mv- is often called the vis viva, but its use is not quite as convenient 
as the definition in the text, and there is a growing tendency towards 
the general adoption of the definition given above. It makes no differ- 
ence, however, which is used, provided it is always used in the same sense. 



68 KINETICS. [114.J 

power in a weight held at a given height is potential; but, 
in descending, it may be made to turn machinery and thus 
do work, in doing which the body passes from a high posi- 
tion to a lower one. The power stored in coal, wood, or 
other fuel is potential / but, if the fuel be burned, it may 
generate steam and thus do work, in doing which it is 
changed from the condition of fuel to that of ashes, cinders, 
smoke, etc. The power in gunpowder is latent ; but, if the 
powder be exploded, it will do work, and may be made to 
throw a cannon ball, or rend rocks, or produce motive 
power. The power contained in food is potential; but 
the food, by nourishing animals, and thus imparting 
strength to them, becomes a source of work. The power 
contained in zinc is potential ;' but the zinc, when acted 
upon by acids, becomes active and capable of driving 
electro-magnetic engines. Air compressed and stored in 
a vessel contains potential energy ; but, by acting upon 
suitable machinery as it expands itself, it will do work. 
The power of steam confined in a boiler is potential ; but, 
if the steam be passed through suitable mechanism, it will 
do work. 

The expression, Change of Position, is sufficiently 
comprehensive to express all the changes of condition. 
Potential Energy has been defined as Energy of Position. 
It is represented by the Greek letter U, the initial of 
potential. 

Heat Energy. 

114. General Statement. — It is well known that heat 
may be produced by friction, or rather, according to our 
present knowledge of the subject, the work of overcoming 
friction produces heat. Thus, the boy, by rubbing a 
brass button briskly on his sleeve, soon makes it too hot 



[116,116.] HEAT ENERGY. 69 

for the comfort of his neighbor's hand. Rubbing two 
sticks together makes them warm. One of the earliest 
methods of obtaining heat was by friction. The heat 
produced by the friction of a match, when rubbed on a 
rough surface, ignites the phosphorus, which, by burning, 
so increases the heat as to set the wood on fire. Axle 
bearings in machinery often become so hot from friction 
as to set fire to the oil and wood which surround them. 
In an experiment made by Sir Humphrey Davy in 1799, 
two pieces of ice, rubbed together in vacuo at a tempera- 
ture below 32° F., were melted by the heat developed at 
the surfaces of contact. 

Iron and other substances may be heated by being 
struck rapidly. Compressing air, or other gaseous bodies, 
develops heat. 

115. The Dynamic theory of heat rests upon the 
hypothesis that heat consists of the motion of the mole- 
cules of a body, or is the result of that motion, and that tc 
produce these motions requires a definite amount of 
mechanical energy. This hypothesis is confirmed by the 
experiments of Joule, who produced heat in a variety of 
ways — compressing air, compressing gases, agitating water, 
and by friction between cast-iron surfaces. 

According to Joule's experiments, if a body weighing 
one pound were permitted to fall freely 772 feet in a 
vacuum, and all the energy thus acquired could be utilized 
in heating one pound of water, it would raise the tempera- 
ture 1° Fahrenheit. This is Joule's equivalent, and in the 
mathematical theory of heat is represented by J. It is 
recognized as the mechanical equivalent of heat. 

116. The Mechanical Equivalent of Heat.— There is 
a definite relation between the work expended in over- 
coming friction and the heat which is produced by it* 



70 KINETICS. [lift.] 

The earliest experiments of which we have any historical 
knowledge tending to prove this fact were made by Count 
Rumford. In 1798 he published in the Trans, of the Royal 
Phil. Society some of his experiments upon this subject. 
A brass cannon weighing 113 pounds was revolved hori- 
zontally, at the rate of 32 revolutions per minute, against 
a blunt steel borer with a pressure of 10,000 pounds. In 
half an hour the temperature of the metal had risen from 
60° to 130° F. This heat would have been sufficient to 
raise the temperature of five pounds of water from 32° tc 
212°. In another experiment the cannon was placed in a 
vessel of water and friction applied as before. In two 
hours and a half the water actually boiled. The heat 
generated in this ease was calculated by Rumford to be at 
least equal to that given out, during the same time, by the 
burning of nine wax candles, three-quarter inch in dia- 
meter, each weighing 245 grains. 

Fourrier, in the year 1807, gave the laws of the trans- 
mission of heat by radiation and conduction, and laid the 
foundation for the mathematical theory of heat. Sadi 
Carnot, in his work entitled " Reflexions sur la puissance 
motrice du feu," published in 1824, compared the energy 
of heat to that of a fall of water from one level to another. 

This branch of science, however, made little or no prog- 
ress until it was shown that there was a definite relation 
between heat and work. Several contemporaneous investi- 
gators entertained the view that such definite relation 
existed, and labored independently to prove it. In 1842 
Dr. Mayer, a physician of Heilbronn, formally stated that 
there exists a connection between heat and work, and first 
introduced the expression, u mechanical equivalent of 
heat" In the same year Colding, of Copenhagen, pub- 
lished experiments on the production of heat by friction, 
from which he concluded that the quantity of heat pro- 



[117.] 



HEAT ENERGY 



71 



duced by friction was directly proportional to the work 
expended. 

But, the most important were the labors of Dr. J. P. 
Joule, of Manchester, England, who, during the years 1840 
to 1843, by a series of very careful and elaborate experi- 
ments, determined a value for the mechanical equivalent 
of heat which is considered as the most reliable ever 
found. (See Phil. Trans., 1850, p. 61.) 

117. Joule's Experiments. — It being impracticable to 
change the energy of a body falling freely into heat in 
such a way as to measure the exact equivalent, Joule 
resorted to different devices. One of the most reliable is 
the following : 

A copper vessel JS, Fig. 30, was filled with water and pro- 
vided with a brass paddle-wheel (shown by dotted lines), 
which could be made to rotate about a vertical axis. The 




paddle had numerous openings, so as to agitate the water 
as much as possible when it rotated in the vessel. The vessel 
was closed so as to prevent the escape of the water, and was 
provided with a thermometer to measure a change of the 
temperature. Two weights, E and F, were attached to 



72 KINETICS. [118.] 

cords which passed over the axles of the pulleys C and ./?, 
and were connected with the axis A, so that as the weight? 
descend the paddle would be made to revolve. The 
height of the fall was indicated by the scales G and H> 
the total fall in Joule's experiments being about 63 feet. 
The roller A was so connected to the axis of the paddle 
that, by removing a pin, the weights could be wound up 
without disturbing the axle. In this way the experiments 
were repeated twenty times. 

The work done by gravity was expressed by {F+ E)xh. 
This was all expended in the following ways: 1st, chiefly 
in overcoming the resistance of the water ; 2d, in over- 
coming the friction of the several bearings; and lastly, 
the kinetic energy in the moving parts at the instant 
the motion was stopped. In the experiments the two 
latter were reduced as much as possible by the mechani- 
cal arrangements, but their effects were computed and 
deducted from the work done by gravity. 

The remaining work was expended in overcoming the 
resistance of the water, and thus developed heat. 

Taking as a unit of heat that necessary to raise the 
temperature of one pound of water 1° F., the unit of heat 

was found to equal approximately 77 r 2 foot-pounds of 
work. More recent experiments by the late Professor 
Roland gave the value, which, is now generally accepted, 
as 778 foot-pounds or 426.8 metre- kilogrammes.* 

If the unit of heat be that necessary to raise one kilo- 
gramme of water 1° C, then a unit of heat equals 424 
kilogramme-metises. 

General Principles of Energy. 
118. Transmutation of Energy. — We have seen that 
kinetic energy may be changed into an equivalent of heat 
* See author's Thermodynamics, page 27. 



[118.] PRINCIPLES OF ENERGY. 73 

energy through the agency of friction. "We also know 
that heat energy may be changed into kinetic energy, as 
is constantly done in the ordinary steam-engine. Other 
energies are constantly brought into action, such as elec- 
tricity, magnetism, chemical action, forces of polarity, and 
indeed any agency by which matter is moved. The rela- 
tions between the different kinds of energies are not well 
known, except that between heat and visible energy ; but 
it is believed that they all consist of some kind of mole- 
cular motion, and are kinetic or potential according to 
circumstances, and that they are changeable one into the 
other. 

This transmutation is constantly going on. To illus- 
trate it with an example : The energy of the sun's heat is 
stored in the plant and there becomes potential. The 
plant may be changed to coal, imparting some of its 
energy to surrounding objects, but concentrating its 
remaining potential energy into a smaller space. Coal is 
raised from its bed by means of kinetic energies, and used 
in a locomotive engine, for instance, where it is burned 
and becomes kinetic. One portion of this energy becomes 
stored in the steam in the boiler ; another portion escapes 
with the smoke ; still another portion escapes through the 
walls of the fire-box ; and another is thrown away with 
the live coals which fall through the grate, or are hauled 
out of the door of the fire-box. 

The steam from the boiler is admitted into the cylinder 
of the engine, and a portion of this energy is utilized in 
driving the piston, and another portion escapes with the 
exhaust steam. The energy of the piston is expended in 
producing heat on the track, heating the axles under the 
cars, overcoming the resistance of the air, wearing the 
couplings between the cars and the working parts of the 
4 



74 KINETICS. [119,120.] 

machinery, bruising the ends of the rails, crushing the 
ties under the track, disturbing the earth or other material 
which forms the roadway, and imparting kinetic energy to 
the train. The energies which are transmitted to these 
several elements finally disappear in heat, which is quickly 
absorbed by the atmosphere, beyond which we are unable 
to trace it with any degree of certainty. 

119. Conservation of Energy. — These principles are 
included in a general law called the Conservation of 
Energy, which may be stated as follows : 

1. The total amount of energy in the Universe is con- 
stant. 

2. The various forms of energy may he converted the 
one into the other. 

It follows from the former of these that no energy is 
ever lost. Energy is indestructible. 

This law is the result of a long series of observations, 
experiments, and generalizations, but it is now considered 
as firmly established as any law in nature. It is accepted 
as a fundamental law in Physical Science, and is as uni- 
versal in its application as the law of gravitation. 

120. Non-equilibrium of Energies. — Energies work 
only as they pass from one condition to another, and this 
is done only when they are not in equilibrium. Thus, if 
two metals are equally hot, one cannot increase the heat of 
the other. A body by losing heat loses energy. When 
steam works by expansion, the temperature is reduced. 
The energy stored in coal is developed by being burned, 
but the heat thus produced exceeds that which it imparts 
to steam or to other bodies. We know of no means by 
which heat energy can be changed into an equivalent of 
kinetic energy ; for in every attempt to accomplish it 
there is an apparent loss of energy, some of it becoming 



[121.] PRINCIPLES OF ENERGY. 75 

potential, or assuming an energy of a lower form. This is 
called dispersion of energy. We have, however, seen that 
visible energy is readily changed into heat. Should all 
the energies of the universe finally become changed into 
heat uniformly distributed throughout space, air motion 
would cease, and the universe would become virtually 
dead. Such a result has been predicted by some writers 
upon this subject. 

121. Perpetual Motion. — By perpetual motion, in a 
popular sense, is not meant ceaseless motion, such as we 
see in the earth and other planets, but a machine which, 
when put in motion, will continue in motion indefinitely, 
without the application of additional power. But every 
machine, when in motion, must overcome the resistance of 
the air and the friction of the bearings, and as these 
resistances cannot be entirely removed, or annihilated, it 
necessarily does work, thus consuming the energy imparted 
to it. When all the energy has been consumed the ma- 
chine will stop and remain at rest. A perpetual motion 
machine is an impossibility. In order to be possible it 
must expend more power than is imparted to it ; or in 
other words, it must possess a self-creating power. 

EXAMPLES. 

1. How many foot-pounds of work is stored in a body 

which weighs 25 pounds, and has a velocity of 100 
feet per minute ? 

2. If a body, moving with a velocity of 5 feet per second, 

penetrates the earth 2 feet, how far would the same 
body penetrate it moving with a velocity of 15 feet 
per second, the resistance of che earth being uniform 
along the path of the body in both cases ? 



76 KINETICS. [121.J 

3. If a tram of cars weighs 60 tons, and moves at the rate 

of 40 miles per hour, how far will it move before 
being brought to rest by friction, the friction being 
8 pounds per ton, no allowance being made for the 
resistance of the air ? 

4. A train of cars weighing 200,000 lbs., and moving at 

the rate of 20 miles per hour, is suddenly stopped ; 
if all its energy be utilized in heating water, how 
many pounds of water would be raised in tempera- 
ture from 32° F. to the boiling point, or to 212° F. \ 

5. In Fig. 29, Article 109, if 17=10 lbs., P=4 lbs., 

fi = 0.2, g — 32J, and the weight W is drawn 5 feet 
by the weight P falling the same distance, when the 
latter strikes the ground ; how far will the weight 
IT move before being brought to rest by friction % 

6. If a piece of iron whose weight is 200 lbs. is moved 

at a uniform rate to and fro on another piece of 
iron, the coefficient of friction between them being 
0.2, what must be the velocity of the moving body 
so that the heat developed by the friction would, if 
entirely utilized, raise the temperature of 5 lbs. of 
water 50° F. every 3 minutes. 



EXERCISES. 

1. Is force the same as energy ? 

2. Does kinetic energy mean work done, or ability to do work ? 

8. An animal eats food which, mechanically speaking, is potential 
energy ; indicate some of the energies which follow the digestion 
of the food. 

4. Why is not wood heated as much by a ball fired into it as iron is by 

a ball fired against it ? 

5, Will a ball flying through the air be warmed on account of the fric- 

tion of the air ? 



[121.] EXERCISES. 77 

6. What becomes of the energy due to the motion of the water in 

rivers ? 

7. Why do a person's hands become warm by rubbing them against one 

another ? 

8. If a ton of coal, which costs 5 dollars, will evaporate a certain 

amount of water, and it requires two cords of wood to evaporate 
the same amount, what will the wood be worth per cord for the 
same purposo f 



CHAPTEK V. 

MOMENTUM. 

122. The equation of Article 86 is 

F=Mf. 

Equation (1) of Article 24 is 

v=ft; 

and, eliminating/* from these equations, gives 

Ft = Mv. 

The quantity Mv is called the momentum of a body 
whose mass is M. This, however, is merely giving a 
name to an expression, but by comparing it with the h'rst 
member of the equation we see that it is the effect which 
a constant force F produces in a time t. We therefore 
call it a time-effect, and represent its value by Q. 

.-. Q = Mv. . . (1) 

If a body has a velocity v when the force begins to act, 
and a velocity v after a time t, then 

Q = Ft = M(v-v ). . . (2) 

123. Momentum of a variable force. — If the fcrco 
be variable, suppose that the time is divided into such 
small portions that the force may be considered as con- 
stant during each portion, then will the total effect be the 
sum of all the elementary momenta. 



1124-126.] MOMENTUM. 79 

Let 2F = an element of time, 

F= the force during any element of time ; 
then 

Q = 2F.At=M(v-v ). 

In these equations F is the effective moving force. 

124. Impulse. — An impulse is the time effect of a blow. 
When one body strikes another, as a hammer striking an 
anvil, its effect is an impulse. The body struck may be 
fixed or free to move ; but the problems which are usually 
considered under this head generally pertain to those in 
which the body considered is free to move. Although the 
effect will be produced in an exceedingly short time, yet 
the result is a time-effect, and is measured in the same way 
as any other time-effect. In the case of a blow, the pres- 
sure between the bodies will be variable during contact ; 
but for the sake of making a practical formula, let 

F'= the mean pressure between the bodies, 
M = the mass of the body considered, 
v = the velocity produced by the blow ; 
then 

Q = F't = Mv. 

125. Momentum is not a force. — The force F is only 
one of the elements of the expression. The unit of mo- 
mentum is the momentum of one pound of mass moving 
with a velocity of one foot per second. 

Momentum is sometimes called quantity of motion and 
sometimes quantity of velocity, but neither fully expresses 
its meaning The expression time-effect is partly descrip- 
tive of its meaning, and is preferred to either of the 
others. 

126. Instantaneous Force. — Wheu an effect is pro- 
duced in an imperceptibly short time, the agency which 



80 KINETICS. [127, 128.J 

produces it is sometimes called an instantaneous force, a 
term which implies that the effect is produced instantly, 
requiring no time for its action. JVb force produces its 
effect instantly, and hence the term is liable to mislead. 
Sometimes it is called an impulsive force, but this is also 
objectionable, for it implies that the effect is a forcej 
whereas the effect is either momentum or work. On 
account of these objections it appears advisable to use the 
term impulse instead of either of the above. 

127. Problems. — 1. If a body, whose weight is 25 lbs , 
is drawn along a horizontal plane by a constant pidl of 
6 lbs., the coefficient of friction being ^, what will be the 
velocity at the end of t seconds f 

The frictional resistance will be 

T Vof 25 lbs. = 2.5 lbs.; 

hence, the effective pulling force will be 

F=6- 2.5 = 3.5 lbs. 

Ft 3.5x193, „. 
•'' V = M == 25x6 = Htfoper sec. 

2. Required the constant force necessary to impart to a 
body, whose weight is 100 lbs., a velocity v during 5 
seconds. 

^= ^? = siSr = *" nearl y- 

Impact. 

128. Impact is the impinging of one body against 
another. 

It is direct when the line of motion of the impinging 
body is normal to the body struck, as in Figs. 31 and 32. 



r 129.] 



ELASTICITY. 



81 



It is central when the line of motion passes through the 
centre of the body struck, as in Figs. 31 and 33. 

It is direct and central when the line of motion pagses 
through the centre of the body struck, and is normal to 
the surface at the point of impact, as in Fig. 31. 






Fltf. 32. 



Fig. 33. 



It is oblique when the line of motion is inclined to sur- 
face at the point of contact. 

It is eccentric when the line of motion is normal to the 
surface of the body struck at the point of contact, but 
does not pass through the centre of the body, as in Fig. 32. 



Elasticity. 

129. Elasticity is that property of bodies by which they 
gain, or seek to gain, their original form after they have 
been elongated, compressed, twisted, bent, or distorted in 
any way. In order to discuss problems involving impact, 
it is necessary to know the laws of elasticity. It is found 
by experiment that, if a body be pulled by a force in the 
direction of its length, as in Fig. 34, it will be elongated ; 
and if the elongation be small it will shorten itself when 
the pulling force is removed. It is found that all known 
substances are more or less elastic. Air and gases are 
highly elastic. 

If bodies regain their original form after the force is 
4* 



82 KINETICS. [130.] 

removed, they are called perfectly elastic. If they regain 
only a part of their distortion, they are called imperfectly 
elastic, and if they regain none of their distortion, they aro 
called non-elastic. The perfect elasticity of solids, even 
within small limits, has been questioned, but for practical 
purposes many of them, such as steel, good iron, glass, 
ivory, etc., may be considered as perfectly elastic. Pris- 
matic bars of good iron and steel may be elongated about 
^5^ of their length without damaging their elasticity. 

130. Coefficient of Elasticity. — Experiments show 
that the elongation or compression of prismatic bars of a 
solid, within small limits, varies directly as the pulling or 
pushing force, and inversely as the transverse section. 

The coefficient of elasticity is the pulling force per unit 
of section divided by the elongation per unit of length. 



B C 



Fig. 34. 



If 1 = AB = the original length of the piece, 
X = BO = the elongation, 
F= the pulling force, 
K = the transverse section, 
E = the coefficient of elasticity ; 
then 

-^r = the strain on a unit of section, and 

XL 

-j = the elongation per unit of length ; 
and, according to the above definition, we have 

F \ Fl . , w 

-=• -$- j = -jjr = constant = A 



[131,132.] ELASTICITY. $3 

Values of E in pounds per square inch. 

For wrought iron from 23,000,000 lbs. to 28,000,000 lbs. 
For steel from 25,000,000 lbs. to 31,000,000 lbs. 

For wood from 1,000,000 lbs. to 2,000,000 lbs. 

131. Elongation of a prismatic bar. — Solving the 
* equation in the preceding article, we have 

X-- ^ 

K - EK> 

which is the expression sought. This expression is true 
only for such strains as do not damage the elasticity. 

132. Modulus (or Coefficient) of Restitution.— If an 
elastic body impinge upon another, the bodies will at first 
compress one another, the compression increasing rapidly 
for a very short time until a maximum is reached ; after 
which, by virtue of their elasticities, they tend to regain 
their original form, and thus force themselves apart. The 
force which causes them to separate can act only while 
they are in contact with one another, but they may con- 
tinue to regain their form after separation. 

The force between the bodies during compression, and 
also during restitution, is constantly changing, and the law 
of change may be very complex. The effect is to constantly, 
but very rapidly, change the velocity of one or both the 
bodies during contact. If the bodies are composed of the 
same substance, and the impact is not so severe as to 
damage their elasticity, the velocity gained by a body dur- 
ing restitution, divided by the velocity lost during com- 
pression, will be constant, and is called the modulus of 
restitution, or index of elasticity. It is represented by e. 
If the bodies are moving in the same direction, the body 
struck will gain velocity during compression, and still 



84 KINETICS. [132.] 

more velocity during restitution. In this case the modulus 
of restitution will be found by dividing the velocity 
gained during the restitution by the velocity gained during 
the compression. In short, it is the ratio of the effect on 
the velocity due to restitution to that due to compression 
According to these principles a simple expression may 
be found for e by assuming that the body struck is at 
rest, and so large that it will be unaffected by a blow from 
the moving body ; for, in this case, the velocity at the in- 
stant of greatest compression will be zero, and hence the 
velocity lost during compression will equal the velocity of 
approach of the striking body, and the velocity regained 
will be that with which it rebounds. 

Let 

v = the velocity of the impinging body at the in- 
stant impact begins ; 
v t = the velocity at the end of the impact ; 
then, 

e = V -l .... (1) 

Let the moving ho&y fall upon the immovable one, and 

let 

H = the height of the fall, 
h = the height of the rebound ; 
then 

= V2gE, Vi = V2gh, 

,...*= v|. . . (2) 

In this way the values of e have been found for the fol- 
lowing substances : 



r iai.1 



IMPACT. 



85 



Substance. 


Value of e. 


Substance. 


Value of e. 


Glass 


0.94 
0.89 
0.81 
0.79 
0.79 
0.73 


Steel, soft 

Cork 


0.67 


Hard baked clay. . . . 

Ivory 

Limestone 


0.65 


Brass 

Lead. 


0.41 
0.20 


Steel, hardened 

Cast-iron. 


Clay, just yielding 
to the hand 


0.17 







For perfectly non- elastic bodies e is zero. If the resti- 
tution were perfect during contact e would be unity, and 
the bodies would be called perfectly elastic. The momen- 
tum lost by a body during compression is generally called 
the force of compression, and that gained during restitu- 
tion the force of restitution. 
133. Problem, — Given the masses and velocities of two 
perfectly free non-elastic bodies before 
impact, to find their velocities after 
impact. 

Take the simplest case, that of direct 
central impact. Let the body Q im- 
pinge upon Q\ the motion being in 
the same direction. 

Let 

M= the mass of Q, 
v = the velocity of Q before impact, 
M' and v' the corresponding quantities for Q f , and 
V = the common velocity after impact. 

The momentum lost by Q during impact will be 

Q = M(v- V); 

and that gained by Q' will be 




Fig. 35. 



86 KINETICS. [134.] 

Q> = M'(V-v'); 
which, being time-effects, will equal one another ; hence, 
M(y- V) = M'{V-v') 

Mv + M'v' 
'"' M + M' ' * I' 

Clearing of fractions gives 

(M+ M')V=Mv + M'v'; . . (2) 

chat is, the momentum of the bodies after impact equals 

the sum of momenta before impact. 

To find the velocity in terras of the weights, substitute 

W W 

— for M and — for M', in equation (1), and we have 

if «7 

1T _ Wv + yy ^ 

y - w+ w ' * • w 

If the bodies are moving toward one another before 
impact one of the velocities will be negative. Making v' 
negative, we have 

Ift-JTV ', 
Jf + Jf ' ' ' * ' w 

in which if J^y = M'v', we have F = ; that is, if two 
inelastic bodies of equal momenta impinge directly upon 
one another from opposite directions they will be brought 
to rest by the impact. 

134. Loss of velocity. — From equation (1) of the 
preceding article we find, by subtracting both members 
fromw 



[135.] IMPACT. 8Y 

and similarly, subtracting both members from v' gives, 

M 
V '- V =-M+M-' {V - V ' ) > • • ® 

which, being negative, indicates that there is a gain of 
velocity. 
135. Impact of perfectly elastic bodies. — It has 

been observed in Article 132 that at the instant of greatest 
compression the bodies have a common velocity ; hence, 
at that instant, the velocity will be given by the equations 
of Article 133 ; and the loss of velocity up to that instant 
will be given by the equations in Article 134. But if the 
bodies are perfectly elastic, the effect upon the velocities 
during restitution will ~be exactly the same as during com- 
pression • hence, the final loss of velocity will be double 
that during compression ; therefore, for the striking body, 
it will be 

WTW> ( - V - V ' ); • • • (1) 

and for the body struck it will be 

-XT3P<— «> ' * (2) 

These subtracted from the velocity before impact will 
give the final velocity. 
Let 

v 1 = the velocity of the former body after impact, 

and 
«/ = the velocity of the latter after impact ; 
then, 



88 KINETICS. [136.J 

9 M 

and these may be reduced to 

Mv + M'v' M' 



>\ — 



M+M' -Jf + WV-^' (5) 



, Mv + M'v' , M . ,, ,„. 
Vl = -MTM^ + M-+M-' {V - V)i (6) 

which, by means of equation (1) of Article 133, become 

*->-3mp<— •■>* • • en 

"'- F + lT¥' ( *- 9 ' ); • • (8) 

which show, as they should, that the final velocity equals 
the common velocity at the instant of greatest compres- 
sion, increased or diminished, as the case may be, by the 
velocity due to restitution. 

136. Discussions of Equations (3) and (4) of the pre- 
ceding article. 

1st. Let M = M', then we have 

Vi = v — (v — v') = v' ; 
Vi = v'+ (v — v') = v ; 

that is, they will interchange velocities. 
2d. Let v' = and M = M r , then 

v t = 0; 

that is, the first body will be brought to rest and the 
second will take the velocity which the first had. 



[137.] ELASTICITY. 89 

3d. Let the bodies be of equal masses, and move in oppo- 
site directions, then M = M' and v' will be negative, and 
we have 

v t = - v f ; 

Vi = v. 

137. Impact of Imperfectly Elastic Bodies. — When 
the bodies are imperfectly elastic the force of restitution 
will be only the 6 th part of the force of compression; 
hence, the velocity due to restitution will be the e th part 
of expressions (1) and (2) of Article 134, or 

eM' . /x 

. eM , n 
-M+-W AV - V) > 

and these, subtracted from the common velocity at the in- 
stant of greatest compression, give the final velocity. The 
results will be the same as equations (5) and (6) of Article 
135, after the last terms of those equations have been 
multiplied by e ; hence, we have 

Mv + M'v' eM' , 

V ^-WVW-WVM' {V -^^ • (1) 

, Mv + 3fv' , eM . 
*= M+M' + MTM' {V - V > ' & 

Let the body M' be indefinitely large and at rest, then, 
•ma in Tig M' = oc and v' = 0, we have 

v x = — ev ; 
t*' = 0; 



90 KINETICS. f 138.] 

hence, the former body will rebound, and by disregarding 
the signs, using the numerical values only, we have 

v ' 

which is the same as equation (1) of Article 132. 

Multiplying equation (1) by M and (2) by M\ and 
adding the results, gives 

Mv x + M'v\ = Mv + M'v' ; 
and, since the index of elasticity has disappeared, the 
total momentum of the bodies before impact will be the 
Same as after impact ; or, in other words, the total momen- 
tum of a free system remains constant. 

138. Loss of Kinetic Energy due to Impact. — The 
total kinetic energy of both bodies, before impact, will be 

iMtf + iM'v*; 
and is independent of the directions of the movement of 
the bodies. After impact the kinetic energy will be 

iMvf+iM'v'f', 
in which substitute the values of v x and v\ from the pre- 
ceding article, and we find 

Mv? + M'v\* = M<tf + M V 2 - (1 7f^ r M \v-v'f. 

1 l M+M' v ' 

Since e 2 is always less than unity for solid bodies, 1— £ 
will be positive, and the last term must be subtracted from 
the preceding ; hence, practically, in the impact of solid 
bodies there is always loss of kinetic energy. The loss will 
be equivalent to the heat developed by the impact. 

If the restitution were perfect, we would have e == 1 ( 
and the expression would become 

iMv t 2 +iM ' v \* = iMtf+±M'v*: 



[188.] ELASTICITY. 91 

hence, in the impact of perfectly elastic oodtes, no energy 
is lost. This result shows the great utility of springs 
under carriages, carts, cars, etc., when they are drawn over 
rough roadways. A horse will do more useful work by 
drawing loads upon a cart, the body of which is supported 
by springs, than if the cart were unprovided with springs, 
and a locomotive will consume less coal in hauling a train 
of cars properly mounted on springs than it would if there 
were no springs under them. 

To find the loss of kinetic energy for perfectly non- 
elastic bodies, make e = in the above equation. 

EXAMPLES. 

1. If a body, whose weight is 20 lbs., is pulled by a con- 

stant force of 5 lbs. for 5 seconds; required the 
momentum produced. 

2. A prismatic bar of iron, whose section is 0.75 of a 

square inch, length 10 feet, coefficient of elasticity 
26,000,000 lbs., is stretched by a pull of 9,000 lbs'., 
what will be the elongation % 

3. A cylindrical bar of iron, whose diameter is \ inch, 

length 2 feet, is elongated -g^ of an inch by a pull of 
2,500 lbs. ; required the coefficient of elasticity. 

4. Two perfectly non-elastic bodies, whose weights are 10 

and 8 lbs., and velocities 12 and 15 feet per second 
respectively, moving in opposite directions, impinge 
upon each other; required their common velocity 
after impact. 

5. A ball, weighing 20 lbs., moving with a velocity of 100 

feet per second, overtakes a ball weighing 50 lbs., 
moving with a velocity of 40 feet per second, their 
modulus of restitution being \ : required their velo- 
cities after impact. Ans. 35-if-, and 65\ feet 



92 KINETICS. [138.J 

6. In the preceding example, suppose the second body to 

be at rest ; required the velocities after impact. 

Arts. — 7|, and + 42-f feet. 

7. In Example 5, suppose that they move in opposite 

directions with the velocities given; required the 
velocities after impact. 

Ans. — 50, and + 20 feet. 

8. A body falls from a height h upon & fixed plane of the 

same substance and rebounds ; the modulus of resti- 
tution being 0, required the whole distance it will 
move in being brought to rest. 

Ans. z 5 h. 

9. In the preceding example find the whole distance when 

e = 1, i, i or 0. 

10. A body impinges upon an equal body at rest ; show 

that the kinetic energy before impact cannot exceed 
twice the kinetic energy of the system after impact. 

EXERCISES. 

1. Is elasticity a force ? 

2. If W is the weight of a body and v its velocity, is Wv the momentum ? 

3. Does momentum enable one to determine the amount of resistance 

which a moving body may overcome ? 

4. Suppose that it is required to determine how far a ball would pene- 

trate a body if fired into it ; would the solution be effected by the 
principles of momentum, work, energy, vis viva, elasticity, or by 
simple force ? 

6. If two bodies move along rough surfaces, and finally impinge upon 

each other, will the formulas of Articles 133 and 135 enable one 
to determine the velocities after impact ? 
6 Can two perfectly non-elastic bodies of unequal masses approach 
each other with such velocities as to destroy their motions ? 

7. Can two perfectly elastic bodies have such relative masses and velo- 

cities that they will mutually destroy each other's motion by the 
impact of one upon the other ? 



[139.] FORCE — ENERGY — WORK, ETC. 93 

8. Is there any relation between the coefficient of elasticity and the 

modulus of restitution f 

9. Explain how the use of springs may prolong the life of cais, and 

also the track on railways. 

Force, Energy, Work, Momentum. 

139. The office which these several elements perform in 
the solution of problems is best shown by an example. 
Suppose that two imperfectly elastic bodies impinge one 
upon the other. That which gives them motion is force. 
That which determines their velocity after impact is 
momentum. That which determines their ability to com- 
press, or break, or damage each other is energy, and the 
compression is work done. The permanent compression 
remaining in the bodies represents kinetic energy lost, 
which has passed into an equivalent amount of heat. 

[The subject of kinetics is resumed on page 202.] 



CHAPTEK VL 

COMPOSITION AND RESOLUTION OF PRESSURES. 

140. Remark. — A force which acts upon a body with 
out producing motion results in pressure only. If several 
pressures concur they may evidently be replaced by a 
single pressure which will produce the same effect as the 
combined effort of all the pressures. The single pressure 
is called the resultant pressure. It might, perhaps, be 
assumed that the value of the resultant for statical pressures 
is the same as for those pressures which produce motion, 
but many think that it is best to deduce the value for the 
former independently of the latter, as has been done in the 
following articles. We will find, however, that the result- 
ant for mere pressures is the same as given in Arts. 52 to 58. 

141. If a pressure acts directly opposed to the resultant 
of all the other forces in the system, and of the same inten- 
sity as the resultant, the system will be in equilibrium. The 
resultant of an equilibrated system is zero. Since, in stati- 
cal problems, there is always equilibrium, any one of the 
forces reversed will he the resultant of all the others. 

142. Component Pressures. — In Fig. 36, if R is the 
resultant of the pressures F and F x , 
the latter are called component pres- 
sures. If the resultant is zero, all the 
forces may be considered as compo- 
nents. There may be any number of 

Fio. 88. "~ components or a single resultant acting 

upon a particle. 

143. If two equal pressures act upon a particle, the 




[144-146.J COMPOSITION OF PRESSURES. 



95 



direction of the resultant will bisect the angle between 
them. 

For no reason can be assigned why it should be neare* 
one than the other of the components. 

144. If three equal pressures act upon a particle, making 
ingles of 120° with each other, they will be in equilibrium. 

For no reason can be assigned why any one should 
prevail over the other two. 

145. Parallelogram of Pressures. — If two pressures, 
acting on a particle, be represented in magnitude and 
direction by two straight lines drawn from the particle, 
and a parallelogram be constructed on these lines as 
adjacent sides, the resultant pressure will be represented 
in magnitude and direction by that diagonal of the par- 
allelogram which passes through the particle. 

The proof of this proposition is given in two parts : — 
first, the direction of the resultant, and second, its magni- 
tude. The following is Duchayla's proof : 

146. Direction of the Resultant of Two Pressures. 
— First, let the forces be commensurable. 




Fia. 37. 




Let F and P be two forces acting on a particle at A ; 
and Ac, Fig. 38, represent F and Ah represent P. Let 
Aa be the common measure of the forces; F= 3Aa, and 
P = 2Aa = 2Ad. On Aa and Ad construct a parallelo- 
gram, which, in this case, will be a rhombus, and the 
direction of the resultant will be the diagonal Ae, for it 



96 STATICS. [147.1 

bisects the angle dAa. Since a force may be considered 
as acting at any point in its line of action, the resultant 
may be considered as acting at e, and the force Aa be- 
comes transferred to de, parallel to Aa, and Ad to the 
line ae. Combining the forces de and dh in the same 
manner, their resultant will be along the line di, and the 
three forces, Aa, Ad, dh, may be considered as acting at 
i ; hence, Ai is the direction of the resultant of these 
forces. Now, combining ae (one-half of ai) with ab, 
gives af; and ef with ei gives ej, and the four forces, Aa, 
ab, Ad, dh, become transferred to^' ; hence, Aj is the 
direction of their resultant. Proceeding in this way, we 
find the resultant of Ac and Ah to be in the direction of 
AB, the diagonal of the parallelogram AcBh. Similarly, 
the proposition will be true for mP and nF, in which n 
and n are integers. 

Secondly, let the forces be incommensurable. A ratio 
may always be found which shall differ from the true 
ratio by less than any assignable quantity, and hence the 
diagonal AB of the commensurable part will differ from 
the direction of the resultant by less than any assignable 
quantity. But no reason can be assigned why this diago- 
nal should be on one side of the resultant rather than on 
the other ; hence, they will coincide. 

147. The magnitude cf the resultant equals the length 

of the diagonal of the par- 
allelograin, of which the 
adjacent sides represent the 
forces. 

Let AB and AG repre- 
sent in magnitude and 
direction the respective 
forces, and AD the direction of their resultant. Take 




[148,140.] COMPOSITION OF PRESSURES. 97 

AE on AD produced backward, and of such length as to 
represent the magnitude of the resultant on the same 
scale as the forces P and F. Then, the forces AB, A 6- 
and A E will equilibrate. On AE and AB, as adjacent 
6ides, construct the parallelogram ABGE, and the diago- 
nal AG will be the direction of the resultant of AE and 
AB. 

Hence, A C must be in the same straight line as A G, 
and A GBD will be a parallelogram ; therefore AD = GB. 
But BG — AE\ .\AE= AD; hence, the resultant of 
AC and AB equals AD in magnitude. 

In Fig. 39 we have 

AD* = AB 2 + AC 2 + 2AB . AC cos BAG, 
or 

B? = F 2 + P 2 + 2FPcos (F,P). 

This equation is of the same form as those in Articles 
14 and 56. 

148. Triangle of Pressures. — If three concurrent 

forces are in equilibrium they may be represented in mag- 
nitude and direction of action by the three sides of a 
triangle taken in their order. 

In Fig. 39, if EA is a force equal and opposite to the 
resultant of the forces P and F, and all three act upon a 
particle at A, they will be in equilibrium ; and, according 
to the preceding Article, if GA represent P in magni- 
tude, and the direction be from G towards A, and AB 
represent Fin magnitude and direction, then will BG 
represent the equilibrating force AE. 

It should be specially noticed that the sides of tha 
triangle do not represent the position of the forces. 

149. Conversely, if three forces, acting upon a parti" 
cle, are represented in magnitude and direction by the 

5 



98 STATICS. [150-152.] 

three sides of a triangle, taken in order, they will keep 
the particle in equilibrium. 

150. If three forces, acting upon a particle, keep it in 

equilibrium, they will be pro- 
c^ F portional to the sines of the 
R ^^ \ angles between the other two. 

< <T' -^ Thus, from Fig. 40, if the 

>v forces P, F, R, acting on a 

& particle at A, keep it in equili- 

FlG. 40. r • i 

brium, we have 
P:F:R = AC: CB : AB, 

= sin ABC : sin CAB : sin A CB, 
= sin (F, R) : sin (P, R) : sin (F, F). 

151. Proposition. — If the directions of three forces %n 
equilibrium are given, and the magnitude of one is also 
given, the magnitudes of the other two may be found. 

The truth of this follows directly from the triangle of 
forces. 

Generally, if three forces, P, F, R, acting on a par- 
ticle, Jceep it in equilibrium, if any three of the quantities 
P, F. R, and the angles which they make with each othe? 
are given, the remaining three quantities may be found, 
provided one of the given quantities is a force. 

For, the solution consists simply in solving a plane tri- 
angle, in which the given parts are a side and any two of 
the remaining parts of the triangle. 

152. If three forces, acting on a particle, keep it in equi- 
librium, they will be proportional respectively to the sides 
of a triangle formed by drawing lines perpendicular to 
the directions of the action-lines of the forces. 

For, a triangle thus formed wilt be simil/ir to the tri- 
angle of equilibrium. 



[153,154.] COMPOSITION OF PRESSURES. 99 

153. If three forces, acting in one jjlane upon a rigid 
body, keep it in equilibrium, their action lines either all 
meet at a point, or are all parallel. 

The lines of action of two of the forces maymeet in 
a point, and their resultant must pass through that point 
and may replace the forces; but, since there is equili- 
brium, this resultant must be equal and opposite to the 
third force ; hence, the line of action of the third force 
must pass through the intersection of the lines of action of 
the other two. If they do not meet they are parallel. 

154. The polygon of pressures and the parallelopi* 
pedon of pressures follow directly from the triangle of 
pressures, giving expressions similar to the corresponding 
proportions for velocity, as in Articles 16 and 17. 



EXAMPLES. 

1. If the angle between two forces is right, what is the 

value of their resultant ? If it is 0° ? If 180° ? 

2. If the forces are 3 lbs., 4 lbs. and 5 lbs. respectively, 

and are in equilibrium, required the angle between 
the forces 3 and 4. 

3. What is the angle between the forces when P=F=R1 

4. If P = F= 100 lbs., and = 60°, find R. 

5. If R = P + F, required 0. 

6. What will be when ~R = P - F% 

7. If P=50 lbs., the angle (P, F)=So° and (P, R)=115°, 

what will be the angle (F, R), and the values of the 
forces F and R for equilibrium ? 

8. A string 7 feet long has its ends fastened at two points 

in a horizontal line 5 feet apart ; a weight of 20 lbs. 
is suspended at a point 3 feet from one end ; required 
the tension on the two parts of the string. 
LoFC. 



100 STATICS. [155, 156.1 

9. Two forces represented by two chords of a semicircle 
passing from a point, show that their resultant will 
be represented by that diameter of the circle which 
passes through the point. 

EXEEOISES. 

1. In Fig. 39, the diagonal joining G and B represents the resultant of 

what forces ? 

2. In the same figure what will represent the resultant of R and F ? 

3. In the same figure, if AD represents a force acting away from A, and 

another equal force should act along the line BG in the opposite 
direction, would they be in equilibrium ? 

4. Under what conditions will two forces be in equilibrium ? 

5. Can a particle be kept at rest by three forces whose magnitudes are 

as 4, 5, and 9 ? Or as 3, 4, and 8 ? 

6. If R is the resultant of P and F, will P and F act when R is acting ? 

Resolution of Forces. 

155. The resolution of forces consists in finding two or 
more components whose united action will equal tfiat of 
the given force. 

156. Rectangular Components. — Let R be a force 
whose components F and P form a right angle with each 
other. Let 

a = the angle between R and P, 
R /S = the angle between R &11&F; 

then we have 

P = R cos a ; 
>P F =R cos/3 = R$ina; 

FI °' 41, R* = P 2 +F*; 

the last of which may be found by squaring and adding 
the two former ; or, by observing that R is represented by 
the hypothenuse of a right-angled triangle, of which the 
sides represent P and F 




[157.] 



RESOLUTION OF FORCES. 



101 



157. To find the angles a and /3. — Since forces may 
act at all possible angles, and be applied at points any- 
where in space, it is desirable to have a definite rule for 
determining the angles which they make with the axes 
of x and y. 

In the first place, draw a line away from the origin of 





Fie. 42. 



Fig. 43. 



coordinates 0, parallel to the line of action of the force, 
and in the direction of action of the force. If the force 
appears to act towards the origin as A 0, Fig. 43, it must 
be prolonged so that it may be represented by OK 

Then, secondly, conceive that the angle a is generated 
by a line starting from the axis 
OX, and revolving about O to the 
left, until it coincides with the 
action-line of the force ; the angle 
thas generated will be + a. In a 
similar way, + j3 will be generated 
by a line revolving from the axis 
OY about 0, to the right. Thus, 
in Fig. 42, a is an acute angle, in 
Fig. 43, it is nearly 360°, and in Fig. 44 it is between 90° 
and 180° ; and /5 in Fig. 42 is acute, in Fig. 43 r between 
90° and 180°, and in Fig. 44, between 270° and 360°. It 
is, however, often more convenient to measure the angle 




Fig. 44. 




102 STATICS. r 158, 159.J 

negatively ; thus, in Fig. 43, — a is the angle XOF, and, 
in Fig. 44, — is YOF 

These rules are arbitrary, but a rigid observance of them 
secures uniformity in practice. 

158. Problem. — To find the rectan- 
gular components of any number of 
concurrent forces in a plane. 

Let O be the point of concurrent 
action, and through this point draw 
two lines, OX and OY, at right an- 
gles with each other. These lines, in 
Analytical Geometry, are called rec- 
tangular axes. 

Let F x , F^ F z , etc., represent the intensities of the re- 
spective forces ; 

a i5 «2> °sj etc., the angles which the forces make 
respectively with the axis of x ; 

Aj A- A», etc., the corresponding angle with the axis 

of y\ 

X, the sum of the components along the axis of x, and 
IT, the sum of the components, along y ; 

then, according to Article 156, we readily find 

X = F x cos Ox + F 2 cos a 2 + F 3 cos a% + etc. = 2 Fcos a ; 
T— F t cosft + F 2 cos/3 2 + F & cos/3 s + etc. = SFcosfi; 

in which the expression "SFooi&a means the sum of a 
series of terms of the form i^cos a. 

It is not necessary that the origin of coordinates be at 
the particle on which the forces act. 

159. Resultant of any number of concurrent forces. 

Let R be the resultant, then, according to Article 156, 
we have 

IP = X 2 + T\ 



160, 161.J RESOLUTION OF FORCES, 



103 



If the given forces are in equilibrium among themselves 

we have 

R = .\ X=0and Y=0. 

160. To find the direction of the resultant we readily 
deduce from Fig. 41, 

X Y 

cos (R, X) = -tj ; cos (i?, Y) = -^ . 



EXAMPLES. 

1. Find the resultant of the concurrent forces in the plane 

xy ; F t = 20, a, = 30° ; F 2 = 30, ^=90° ; ^=40, 
og == 150° ; and F 4 = 50, a 4 = 180°, and find the 
angle between R and x. 

2. If the forces F x = 20, ^ = 180° ; i^ = 10, a 2 = 270°, 

are concurrent, find R. 

3. If four forces are all equal to each other and concur- 

rent, and «! = 0, 02 = 90°, a 3 = 225°, a, = 270°, find 
R and the angle which it makes with the axis of x. 



161. If the forces are referred to three 
rectangular axes, we have 

X= Fi cos ai+Fi cos a 2 +etc. = Si^cos a ; 
T= Fi cos /?i + #, cos /? 2 +etc. = XFcos p ; 
Z = Fi cos 71 + i^ cos y a + etc. = Ei^cos y ; 



Y 1 



cosa 



5 ;oo S / 3 = 5 ;co B j= 1 . 



Fig. 49. 



CHAPTEK VH 



MOMENTS OF FORCES. 

162. Definition. — The moment of a force is a measure 
of its effect in producing rotation, or of its tendency tc 
produce rotation. 

163. Measure of the Moment. — The moment of a 
force, in reference to a point, is the product of the force 
into the perpendicular distance of the action-line of the 
force from the point. 

Let a particle w or w\ Fig. 47, be connected with a 



w 

-A 

urn 



>F 




Fig. 47. 



Fig. 48. 



point O by a line without weight, and let a force F act 
at any point A of this line and perpendicularly to it. 
The effect of the force will vary directly as the distance A 
from the point O. Suppose that rotation has been pro- 
duced, as shown in Fig. 48, the particle having been moved 
through an angle A OB. The point of application of the 
force will have moved over the space AB, and the work 
done by the force will be (Art. 92) 

F.AB. 
If the point of application were at A\ it would have 



[164,165.] MOMENTS OF FORCES. 105 

moved over the space A' B r in moving the particle w from 
A to B y and the work done by F would be 

F. A'B'. 
But, 

AB: A'B'::OA: OA'; 
hence 

F.AB:F.A'B'::OA: OA' ; 

that is, the effect of a force, in producing rotation, varies 
directly as the perpendicular distance of the force from the 
point about which rotation takes place. The effect evi- 
dently varies directly as the intensity of the force ; hence, 
generally, the effect will vary as the product of the force 
into the distance of the action -line of the force from the 
point. This is called the moment, as given above. 

164. If 'the line of action of the force is inclined to the 
line OA, resolve the force iuto two compo- 
nents, one F 2 , acting along the line A 0, 
the other, F ly acting perpendicular to OA 
at the point of application A, of the given 
force. The former does not tend to pro- 
duce rotation about O, but the latter acts in 
the same manner as in Fig. 47. Hence, 
we have, for the measure of the moment in w * „ 

? Fio. 49. 

this case, 
F x . OA =F&in OAB .OA = F. OA sin OAB = F. OB. 
But OB is the perpendicular from O upon the line 
AB of the force F. 

165. Axis of Moments. — Eotation about a point is 
always equivalent to a rotation about an axis which passes 
through the point, and perpendicular to the plane of 
motion of the particle. This line is called the axis of rota- 
tion, and, in reference to the moments of forces, is the 
axis of moments, or moment axis. 

5* 




106 STATICS [166-163.1 

166. If the axis of rotation is fixed, and the line of 
action of the force is inclined to the plane of motion; in 
order to find the moment of the force, the force is re- 
solved into two components, one of which is perpendicu- 
lar to the plane of motion and the other parallel to it. 
The former has no moment in reference to the fixed axis 
and the moment of the latter will be found by Article 
164. 

167. The point or axis about which the particle or body 
rotates may not only move in space, but may also change 
its position within a body. 

168. Definitions. — The point O, where the axis of rota- 
tion pierces the plane of motion of the particle, or the 
plane in which it tends to move, is called the origin of 
moments. 

The perpendicular OB, let fall from the origin of mo- 
ments upon the action-line of the force, is called the arm 
of the force. 

The moment of a force in reference to a point is the 
product of the force into its arm. If a is the arm of the 
force the moment will be 

Fa. 

The moment of the velocity is the product of the velo- 
city into the perpendicular from the origin of moments 
upon the direction of motion. The uirection of motion 
will be in a tangent to the path at the point where the 
velocity is considered. Let j> be the perpendicular, then 
the moment of the velocity will be 

pv. 

The moment of the momentum is the continued product 
of the mass, velocity, and perpendicular from the origin of 



[169-171.] MOMENTS OF FOBCES. 107 

moments upon the direction of motion. Let Q be the 
momentum, then 

Qp = Mvp. 

This principle is important in the solution of certain 
problems involving an aggregation of particles. 

The moment of a force oblique to the axis of rotation 
is the product of the component of the force on a plane 
perpendicular to the axis into the arm of the component. 

Generally, the moment of a mechanical agent is a meas- 
ure of its importance in producing, or tending to pro- 
duce, rotation. 

169. The moment of a force may be represented by 
twice the area of a triangle, of which the base 
represents the magnitude and position of the /\ 
force, and whose apex is at the origin of mo- / j \ 
ments. For, the altitude of the triangle will / \ 

be the perpendicular upon the force, and ^ ^ 

hence, will be the arm of the force. FlG ' 50 ' 

170. Sign of the Moment. — If a force tends to turn 
a system one way, it may be considered positive ; then, if 
in the opposite direction, it will be negative. Either 
direction may be considered positive, but when chosen it 
must not be changed in the solution of a problem. 

If a watch be placed at the origin of moments, with its 
face in the plane of the force, the moments of those forces 
which tend to turn the particle or body in the direction of 
the movement of the hands of the watch will be right- 
handed, and in the opposite direction left-handed. 

171. The value of a moment may be represented in 
magnitude and direction by the axis. When the moment 
is positive, let the axis be represented above the plane, as 
in Fig. 51, and a distance laid off on it to represent its 



108 



STATICS. 



[173.] 



magnitude. When the moment is negative, lay off the 
axis below the plane of the force, as in Fig. 52, 





-Fa 



Fig. 51. Fig. 52. 

172. Composition of Moments. — Since moments are 
fully represented by lines, they may be compounded or 
resolved in the same manner as forces. Thus, if the 
forces are all in the same plane, their axes will be paral- 
lel, and, if they have the same origin, their axes will 
coincide ; in which case, the resultant moment will be the 
algebraic sum of the several moments. 

If O r be the resultant moment, we have 

O r = Fa + F x a x + F^ + etc. 

If the forces act in different planes let all their moment 
axes pass through A, Fig. 53, 
and let 

O x = F x a x = the moment of 

one force, 
2 = Fjfa = the moment of 

another force, 
6 = the angle between the mo- 
ment axes, and 
O r = the resultant moment of 
1 and 2 ; 
then, as in Article 147, we have 

O r 2 = 1 2 + 0} + 2<9 1 6> 2 cos 0. 

If the axes do not pass through a common point, it may 




[173,174.] 



MOMENTS OF FORCES. 



109 



still be proved that the resultant moment may be found 
from the preceding equation. 

173. Proposition. — If any number of concurring forces 
are in equilibrium, the algebraic sum of their moments 
will be zero. 

Let F x , F 2 , F z , etc., be the 
forces acting upon a particle at 
A. Take the origin of moments 
at any point O. Draw OA, and 
let fall the perpendiculars Oa, 
Ob, Oc, etc., upon the action-lines 
of the forces ; and let 
Oa = a x ; Ob = a 2 ; Oc = a 3 , etc. 

Since they are in equilibrium, 
the sum of the components of all the forces perpendicular 
to OA (see Article 142) will be zero ; hence, 

F 1 sinOAF 1 + F 2 smOAF 2 + F s &mOAF s + etc. = 0; 
or 




Fig. 64. 



w Oa . w Ob 
Fl OA + F *OA 



+ i^ + etc. = 0. 



or 



Multiplying by OA, we have 

F t .Oa + F 2 .Ob + Fs.Oc + etc. = 0; 

■Fi<h + Fifh + F z a z + etc. = SFa = 0. 

When there is equilibrium in reference to rotation, any 
one of the moments may be considered as equal in value 
but directly opposed to the resultant of the moments of 
all the other forces. 

174. Unit of Moments. — If the force be given in 
pounds, and the arm in feet, the unit will be a foot-pound, 
and the moment will be a certain number of foot-pounds 



110 STATICS. [175-177.] 

of rotary effort. It will be observed that this is Dot the 
same as foot-pounds of work. 

175. The origin of moments may he taken anywhere 
in the plane of the forces. This is evident from the pre- 
ceding article ; for, the point O, in Fig. 54, was not only 
chosen arbitrarily, but no trace of its position remains in 
the result. This is also evident from the fact that if the 
forces are in equilibrium their tendency to turn the body 
about any point is zero. In statical problems, therefore, 
the origin may be chosen arbitrarily. 

176. The arm of a force in terms of Rectangular 
Coordinates. — Let F be a force, of which Aa is its line 

of action. Take the origin of 

coordinates at O, which is also 

taken as the origin of mo- 

X ments ; then Oa will be the 

5£ j arm of the force. 

jr- 9\r *— Take any point A, in the 

B^X/' l me Pf tne force, and drop the 

FlG 31> perpendicular Ah, and from 

its foot drop the perpendicular 
be upon aA, and draw Od parallel to aA. 

Let a = dOh; /3 = the angle between i^and Y = cAh 
y = Ah ; x = Oh. 

Then 

cb = y cos a ; dh — x cos ,6 ; 

.-. aO = ch — dh = x cos ft — y cos a. 

177. The origin of coordinates may be at one place and 
the origin of moments at any other, for the positions of 
both are arbitrary. Thus, in Fig. 55, the origin of mo- 
ments may be taken at d, or h, or any other place, the 
origin of coordinates remaining at 0, 



[178,179.] 



MOMENTS OF FORCES. 



Ill 



178. Moments of Parallel Forces. — If the forces are 
parallel their arms will coincide. Thus, in Fig. 56, if O 
be the origin of moments, then will the arms of the forces 
F x , F 2 , F s , etc., be Oa, Ob, Oc, etc., respectively. 



'*<r 



+ Fs 



a 



•■*•/ 



Pig. 




Fig. 57. 



179. Problems. — 1. A weight W is suspended from one 
end of a horizontal bar A G, the other end of which rests 
against a vertical wall, the bar being held in position by 
a cord DB ; required the tension of the cord. 

Take the origin of moments at A, where the bar touches 
the wall. The perpendicular AE upon the cord BD will 
be the arm of the tensile force of the cord, and A C will 
be the arm of W. Let t be the tension of the cord, then 
will t.AEbe the moment of the tension, and we have 

t.AE=W.AO. 

This problem illustrates the mechanical arrangement of 
the bones and muscles at the elbow-joint, by means of 
which the arm may be held in a horizontal position and 
support a weight in the hand. The joint is at A, the 
hand at C\ and the muscle at DB\ but in the arm the 
distance AE is much less in proportion to A C than that 
shown in the figure. 

2. A weight W is suspended by a cord DB, and pushed 



112 STATICS. [180,181.1 

from a vertical by a bar AB ; required ths tension of 
the cord. 

Take the origin of moments at A, and drop the perpen- 
diculars A C and BE. The perpendicular BE will be 
of the same length as if it were drawn 
2 horizontally from A to a vertical 

.X through B. If t be the tension, we 

/ \ have 

*ST\ ' x t.AC=W.BE 

'" - J^P 

■— "l v BE W 

Fig. 58. ^ v 

If the length and inclination of AB be given, we have 
BE =AB sin 0. 

If the length of DB be also known, the angle ABB = <f> 
may be found. Or, if the three lines AD, AB, DB, are 
given, the angles </>, 0, ABD, may be found by solving the 
triangle. 

180. Choice of the Origin of Moments. — When the 
forces are in one plane, the problem may be solved by a sin- 
gle equation of moments, provided the origin of moments 
can be so taken that the moment of one force only will be 
unknown. It will be observed that the forces at the 
origin of moments have no moments, and hence do not 
enter the equation of moments. When there are several 
unknown forces, it is possible, in many cases, to find the 
value of all of them by taking the origin of moments at 
different places, so as to involve only one unknown quan- 
tity at a time. 

181. Problems. — 1. In Fig. 58, suppose that the bar AB 
rests against a smooth surface DA, and is prevented from 



T181.] MOMENTS OF FORCES. 113 

sliding upward by a string AE ; required the tension of 
the string and the pressure against the surface. 
Let 

F= the tension on AF, and 
X = the pressure against the surface. 

If the origin of moments be taken at B, both the un- 
known forces i^and X will enter the equation, and hence, 
neither will be determined. If it be taken at D the mo- 
ment of F will be zero. The forces F and X are the 
components of the compression along BA ; hence, the 
latter will not be considered when the former are. 
Taking the origin of moments at D, we have 

X.DA = W.FB, 

Now, taking the origin of moments at B, there results 

F.BF=X.AF; 

in which, substitute the value of X from above and reduce, 
and find 

r DA m 

The compression along AB will be 



VI^ = VJE&+A& AB 
■ DA DA 

The compression may be found directly by taking the 
oiigin of moments at D. Drop a perpendicular from D 
upon BA prolonged (which the student can draw in the 



114 



STATICS 



[181.] 



figure), and call it a. Let c = the compression ; then the 
equation of moments will be 

c.a = W.EB; 



but 



a = DA sin 6 
EB 



DA 



EB 
AB* 



c = 



w=iMw, 




Fig. 59. 



a DA 

as before. 

2. Two forces, P and E, act as a pull upon the ends 
of a bar AB, both of which are inclined to the bar at 
known angles, the point O being 
fixed, and the distance OB 
known; required the distance 
A C for equilibrium. 

Take the origin of moments at 
O, and drop the perpendiculars 
Oa and Cb ; then the equality of 
moments gives the equation 

P.Oa = F.Ob. 

The distance Ob is known from the equation 

Ob= OBsmOBb; 

and A O from the equation 

AO= Oa cosec a A O. 

Substitute in this equation the value of Oa found from 
the first equation, and Cb from the second, and we find 

P sin a A O 
The resultant of the forces P and F will pass through 




[181.] MOMENTS OF FORCES. 115 

their intersection D, and also through the fixed point C. 
The moment of the resultant, when the origin is at any 
point upon it, will be zero. Its value, however, may be 
found by taking the origin of moments at A or B. 

3, A weight P x is made to act vertically upward at the 
end of a rigid bar OA, and P 2 vertically downward at B, 

on the same bar, the bar being 
free to turn about the end O ; 
required the distance Ob for 
equilibrium. 

Take the origin of moments 

figM50. " at ®-> an ^ draw the horizontal 

line Oc, cutting the verticals 

through B and A at b and c respectively ; then will the 

equation of moments give 

P 1 .Oc-P 2 .Ob = 0. 

-* 2 

If the distance be between the weights is given, then 
Oc=Ob + be; 
which, in the preceding equation, gives 

EXAMPLES. 

1. In Fig. 57, if W= 20 lbs., AG = 2 ft., AB = 6 in., 

and AD = 4c in., find the tension on the cord DB. 

2. In the preceding example find the horizontal pressure 

against the wall. 

3. In Fig. 57, if the surface at A is perfectly smooth, find 

the vertical force F applied at A, which will just 
prevent sliding. 



116 STATICS. [182.] 

4. If the cord BB, Fig. 57, is inclined 45°, what must be 

the distance from A to B that the tension on the 
string shall equal the weight W ? 

5. In Fig. 58, if BB = 2AB, = 45°, and W= 50 lbs., 

required the tension on the cord and the compres- 
sion on the bar. 

6. A strut BO, free to turn about its lower end, supports 

a weight W from its upper end, 

5jp| the strut being held by a cord A O, 

s'g one end of which is attached to the 

>< g A upper end of the strut and the 

s' g w other to a point A in the horizontal 

-a jb b plane: required the tension of the 

Fig. 61. i 

cord. 

7. In the preceding example, what condition must be ful- 

filled that the tension of the cord will equal the 
weight TF? 

8. In Fig. 61, if W= 500 lbs., AB = 6 ft., BB = 4 ft., 

and DC = 8 ft., what will be the compression upon 
GB% 

9. In Fig. 60, if P x = P % and be = 2 ft., what will be the 

distance Ob for equilibrium ? 

10. In Fig. 60, if Ob = 2 ft., and P x = P 2 , what will be 
the distance be for equilibrium '( 

Couples. 

182. Definition. — Two equal unlike parallel forces, 
whose lines of action are not coincident, form a couple. 
The last equation of the preceding article is 

p 

01)z=z P J p lc 'i 

-L 2 ± 1 



[183,184.] COUPLES. 117 

in which, if P 2 = P ly it becomes 
Ob = 00; 

that is, in order that two such forces shall be in equili- 
brium, in reference to rotation, the origin of moments 
must be at an infinite distance from the forces ; in other 
words, two equal parallel forces, whose directions are con- 
trary and lines of action not coincident, cannot be in 
equilibrium in reference to rotation. Such a system has 
received a special name, called a couple. 

183. The office of a couple is to produce, or to tend 
to produce, rotation only. For, the only other effect which 
it can produce is that of translation ; but, since the forces 
are equal and directions contrary, whatever translation is 
produced by one force in any direction will be exactly 
neutralized by the other in the opposite direction ; hence, 
they cannot produce translation. 

184. Moment of a Couple.— In Fig. 62, let O be the 
origin of moments, and P at A equal P at B, one mo- 
ment being right-handed and the 
other left-handed. The moment 
will be 

P. OB- P. OA (not equal 0). o ~~ 
But, 

OB- OA+AB; fio.68. 

hence, the preceding expression becomes 

P.OA+P.AB- P.OA; 
or, simply 

P.AB; 

that is, the moment of a couple is the product of one of 
the forces into the perpendicular distance between the 
lines (faction of the forces. 

This is independent of the origin of moments. If the 



118 STATICS. [185-187.3 

origin be at A, upon the line of action of one of the forces, 
the moment of the couple will be the same as the moment 
of one of the forces. 

185. A couple can be equilibrated only by an equiva- 
lent couple having a contrary moment. For, the only 
effect being rotation, such a system of forces must be em- 
ployed as will produce a contrary rotation, and this requires 
an equivalent couple. 

186. A resultant couple is one which will produce the 
same effect as the several couples. 

If P l9 jP 2 , etc., be the forces of several couples all in 
one plane, 

a t , a 2 , etc., be their respective arms, 
P, one force of a resultant couple, and 
r, the arm of the resultant couple ; 
then 

Rr = P x a^ -f P 2 a 2 + P s a 3 + etc. 

If the couples are in equilibrium, any one of them may 
be taken as equal but contrary to the resultant of all the 
others. 

187. Proposition. — If two couples, having equal mo- 
ments, but whose directions of action are contrary, act upon 

a body, they will equilibrate each other. 
This is evident from Article 185, but the 
proposition is presented here in order to 
show that the forces constituting the 
couples may be applied at any point of 
the body, and that the arms of the cou- 
ples need not be parallel. 
Conceive that any point O in the body 
is fixed, and taken as the origin of moments, then will the 
moments of the couples in reference to this point be 
P x . ab — P % . cd ; 




[188,189.] COUPLES. 119 

but, since their moments are assumed to equal each other 
we have 

P 1 . ab - P 2 . cd = ; 

hence, there is no tendency to turn about O. The same 
may be shown in reference to any other point of the body ; 
hence, the body will be in equilibrium in reference to 
rotation. 

188. Proposition. — A force, acting at any point of a 
body, is equivalent to an equal parallel force at the origin 
of moments, and a couple whose mo- 
ment is the moment of the original i p 
force. A 

Let the force P be applied at A, 
and the origin of moments be at B. 
At B introduce two equal and oppo- >k & 
site forces, each equal and parallel to 
the original force P. Since the two forces at B neutral- 
ize each other, the effect of the three forces will be the 
same as that of the single force P. But, by making a 
new combination of the forces, we have the force acting 
down at A, combined with the equal parallel force at B 
acting upward, constituting a couple whose arm is AB, 
and the force P at B acting downward. 

When a body is free to move, a single force acting upon 
it may produce both rotation and translation, and it may 
be shown that it will produce both, unless the line of action 
of the force passes through the centre of the mass of the 
body. 

189. Proposition. — When several forces have a result- 
ant, the sum of their moments will be zero when the 
origin of moments is upon the line of the resultant, for 
the moment of the resultant will be zero. 



Fig. 64. 



120 



STATICS 



f 1901 J 



Three Parallel Forces. 

190. The relation between three parallel forces in equi* 
iibrium may be found by means of the principles of mo- 
ments, and the result may be extended to any number of 
parallel forces. 

In Fig. 65, let the forces P, F, P, act in parallel lines 
and be in equilibrium. 



h' 






j^p 



E 



E 



V 



Fig. 65. Fig. 66. 

Taking the origin of moments at D, we have 

F.Z>G = P.DF; 

and if the origin of moments be at F y we have 

F.CE '= R.DF. 

Adding these equations gives 

F{DC+ OF) -= (F + F)DF; 
but, 

DC+ OF=I)F, 

which, in the preceding equation, gives 

F=P + P; 
hence, the force F, acting in one direction, equals the sum 
of the two forces acting in the contrary direction. 
If i^and P are given, we have, by transposition, 

F- P = E. 
The same principles apply to Fig. 66, in which the 
directions of P and F are the reverse of those in Fio:. 65. 



[191, 192.J PARALLEL FORCES. 121 

By transposing R in the last equation we have 
F- P - R = ; 
in which, if F be called the typical force, and the alge- 
braic signs be understood, we may write it 

2F=0; 
and this expression is true for any number of parallel 
forces. 

191. A rigid body, being acted upon by any Dumber of 
parallel forces in one plane, it is necessary and sufficient 
for equilibrium that we have 

2 Fa = ; 

SF=0; 
the former of which will determine equilibrium in refer- 
ence to rotation, and the latter in reference to translation. 

192. A single force and a single couple in one plane are 
equivalent to a single force equal 

and parallel to the original sin- a"-** 

gle force, but having another 
point of application. jd _ 

If they are parallel, as in Fig. 
67, the resultant of the forces in 
reference to translation will be fig. 67. 

F+F-F=F 

Call this resultant F', to distinguish it from the F in 
the figure. A force equal and opposite to F', acting at 
Borne point J), will produce the couple F'—DC—F, which, 
for equilibrium, must be equal and opposite to P—AB— P, 
Hence, to find DO, we have 

F.DO=P.AB 

:.DC=?-AB. 

6 F 



122 STATICS. [193,196.] 

If the forces are not parallel, combine F with P, and 
their resultant with — F, and the same result will be 
obtained. 

i93. Remark. — The principle of moments, strictly 
speaking, is applicable only to problems involving extended 
masses ; for, although we speak of the moment of a force 
in producing the rotation of a particle, yet, in order to 
realize it, it is necessary to assume that the particle is 
connected with the point about which rotation takes place, 
by means of a rigid bar, which is itself a finite body and 
not a particle. 

194. Proposition. — A system of forces acting in one 
plane, if not in equilibritom, must be equivalent to a sin- 
gle force or to a couple. 

For, the resultant of two forces may be found, and the 
resultant of that resultant and a third force, and so on, 
and if their lines of action thus intersect each other a sin- 
gle resultant may be found ; otherwise a single couple 
may be found. 

195. Proposition. — A system of forces in one plane y 
acting on a rigid body will be in equilibrium if the alge- 
braic sum of the moments of the forces vanishes in refer- 
ence to three points in the plane not in a straight line. 

For, if they are not in equilibrium in reference to rota- 
tion, the algebraic sum of the moments could not vanish 
for any point in the plane; and if they had a single 
resultant, then the moments would vanish only for points 
on the line of the resultant. 

196. Problems. — 1. A prism AF is acted upon by a 
couple in the plane of the upper base ; required the value 
of the couple in the plane of the lower base that will 
equilibrate the former. 

The forces in the plane of the upper base tend to turn 



196.] 



PARALLEL FORCES. 



123 



the prism about an axis perpendicular to the base ; simi- 
larly the couple in the lower base will tend to turn it about 
the same axis ; hence, if their moments are equal and 
contrary, they will equilibrate each other. 

This shows that couples of equal moments in paraLel 
planes are equivalent. 

The couples, in this case, twist the body upon which they 
a 





Fig. 69. 

act. This effect is called torsion. The amount of torsion 
depends upon the properties of the material, as well as 
upon the size of the body and the moment of the couple. 
The properties of materials are investigated in works upon 
the Resistance of Materials. 

A single force applied at the end of a lever, as in Fig. 
69, will not only twist the body, but will push it side- 
wise. For, as we have seen in Article 188, it will be equi- 
valent to a couple whose moment is 

P.AB, 

and a force 

p, 

applied at A, the former of which twists the body and the 
latter pushes it sidewise. This may be easily illustrated 
by the student in a variety of ways, such as turning an 
auger by one handle, twisting a long rod by means of a 
single-handed wrench, etc. 



124 STATICS. [196.] 

2. A door, gate, or frame, supported by two hinges, 
carries a weight ; required the pressure upon the hinges. 

When hinges are employed, 
L =J5 — jt they may be so set that one or 

the other will carry all the 



,. =*c vertical pressures. In Fig. 70. 
y^' Q there being no provision for car- 



rying any of the vertical pres- 
sures at the upper bearing, the}' 
\ T will all be supported at the lowei 

Pig. 70. , rr 

end. 
Taking the origin of moments at B, we have 

H.BA = W.DC 

Similarly, taking the origin at A, we have 
R^BA=zW.DG 

.:* = §£*; 

hence, 

and as they are parallel they constitute a couple. The 
only remaining force is the vertical one at B, and is called 
V. This force, combined with W, must constitute the 
equilibrating couple ; hence, 

V=W. 

The total pressure at the lower end is the resultant of 
Fand H ly and hence is 



V 



BA* ~ BA • 



[196.] 



PARALLEL FORCES. 



126 



EXAMPLES. 

1. If three forces are represented in magnitude, position^ 

and direction of action by the sides of a triangle 
taken in their order, show that they are equivalent 
to a couple. 

2. In Fig. 71, if the forces act along the sides of a trian- 

gle, P from B toward O, F from C toward A, and 
M from B toward A ; show that for equilibrium in 
reference to rotation, R = B+ K 

3. Equal weights are suspended at the corners of a trian- 

gle ; required the point where the triangle must be 
supported that there will be equilibrium. 




v f 


D 








W//////%. 







Flo. 71. 



Fig. 72. 



4. Two men carry 175 lbs. between them on a pole, resting 

on one shoulder of each ; the weight is twice as far 
from one as from the other ; how much weight does 
each carry, neglecting the weight of the pole ? 

5. If a prismatic block of stone, whose width AB is 2 ft., 

height AC is 3 ft., weighs 500 lbs., and it be con- 
sidered that the whole weight acts at the centre g, 
what force, acting horizontally at C, will j ast turn 
the block about the edge B % 

6. A man, whose weight is 175 lbs., desires to raise a body 

which weighs 4,000 lbs. by means of a lever 8 feet 
long ; one end of the lever being placed under the 



126 STATICS. [196.J 

body, how far from the end shall the fulcrum be 
placed so that his weight at the other end shall just 
balance the body ? 



EXERCISES. 

1. The force being given in pounds, and the arm in feet, is it proper to 

say — a moment of a certain number of foot-pounds ? Foot- 
pounds of what ? 

2. What is the moaning of foot-pounds of work ? 

3. What is meant by foot-pounds of momentum per second ? 

4. If velocity is given in feet, and the arm also in feet, what will be the 

unit of the moment of momentum ? 

5. Can a single force acting upon a rigid body produce rotation if there 

is not a fixed point in the body ? 

6. Will two couples acting upon a rigid body in planes at right angles 

with each other produce translation ? 

7. If a person supports a weight of 100 lbs. suspended from a rod upon 

his shoulder, and he pulls down upon the rod with a force of 25 lbs. 
with his hands so as to balance the weight, how much more than 
his own weight will be the pressure of his feet upon the earth. ? 

8. If a hole be bored by an auger, is the resistance to the cutting 

equivalent to a couple ? 

9. In Fig. 70, will the pressure H at A be increased if the weight W 

be placed at a greater distance from D ? 



CHAPTER YIIL 

PARALLEL FORCES. 

197. Parallel Forces are such as act along para.lel lines. 
They may be conceived as concurring in a point at an in- 
finite distance, but this would be equivalent to saying that 
they do not actually concur. They are forces not acting 
upon a single particle, but upon the several particles of a 
body. 

198. The Resultant of Parallel Forces.— Let R be 
the resultant of the parallel forces F ly F 2 , etc. ; then ; 
according to Article 191, we have 

B = F 1 +F 2 +F B + etc. = SF ' , * /' 

It will be observed that the value — ^j r 3 

of the resultant is independent of the — ^ 

points of application of the forces. ^ ' 

If the forces are in the plane xy, and 
are resolved parallel to the axes x and y, we have, accord 
ing to Article 156, 

X= (F 1 + F 2 +F 3 + etc.)cosa = cos a2F; 
Y= (F 1 +F 2 +F 3 + etc.) cos/3 = cos 82 F= sin aSF. 
Squaring and adding, gives 

m = X 2 + Y 2 = (XFf (cos 2 a + sin 2 a) 
= (2 F) 2 

which is the same as given above. 



128 



STATICS. 



[199 200.J 



If the given forces are in equilibrium, we have 
R = 0; 
.\SF=Q; X=0; Y=0. 

199. The expression for the moments of parallel forces 
is given in Article 178, and the condition for equilibrium 
in reference to rotation, in Article 191. These expressions 
are independent of the points of application of the forces ; 
but when these points are given the equation of moments 
is not only simplified, but it is found that there is always 
a point on the line of action of the resultant, called the 
centre of parallel forces, which possesses an important 
property. 

200. Centre of Parallel Forses. — The centre of par- 
allel forces is that point through which the resultant will 
constantly pass as the forces are rotated about their points 
of application, the forces remaining constantly parallel as 
they are rotated. 

To illustrate, let the parallel forces P, E, and R be in 
equilibrium, having their points of application at P, A, 
and G in the straight line PC. 
The point G will be the point of 
application of the resultant of P 
and R. Draw GE perpendicular 
to the lines of action of the forces ; 
then, since there is equilibrium, 
we have 




P.EG=R.DC. 



0) 



Conceive that the forces are revolved through an angle 
of 90°, retaining their relative directions of action. The 
force R will then act at A, parallel to EG and to the 
right (or left), and P will act at the point P, also parallel 



IfiOl.J PARALLEL FORCES. 129 

to EG and to the left (or right). In the new position the 
arm of P, in reference to (7, will be DA, and of P, will 
be EP ; hence, if there is equilibrium, we will have 

P.EP = B.DA. . . . (2) 
For, from the similar triangles OP A and CEP, we have 

EG:PO::EP:PA. 
~n a 

which, substituted in equation (1), gives 

P. EP = P.DA; 

which is the same as equation (2) ; hence, the resultant in 
the new position will pass through O. 

In a similar way it may be shown that it will pass 
through G for any amount of rotation of the forces P and 
P) hence ; G is the centre of the parallel forces P 
andi?. 

Similarly, A is the centre of the parallel forces F and 
P, if G and P are the points of application of the re- 
spective forces. But if E be the point of application of 
P, and G of E] then A will not be the centre of those 
forces. 

The centre of two parallel forces will be on the line 
joining their points of application ; and by combining 
their resultant with a third force in a similar way, the 
centre of three forces may be found, and so on for any 
number of forces. 

201. To find the Centre of any Number of Paral- 
lel Forces. — Let the forces be in the plane xy, their 
points of application being at A, C, E, etc. The centre 



130 



STATICS. 



[301.1 



of the forces F[ and F 2 will be at B, where their resultant 
intersects the line A O. Let 

Xx, y x be the coordinates of -4, 



and 



•'.y 
«",«" 






x, y be the coordinates of the centre of all 
the forces, which, being on the result- 
ant, is called the point of application 
of the resultant. 




Draw Aq and Br parallel to OX, Am, Bn, etc., parallel 
to OY, and Oe perpendicular to the lines of action of the 
forces. The triangles AqB and Br C are similar, and give 
AB:BC::Aq:Br 

: : On — Om : Op — On 
::x' — a^: Xz — x'. 



[202.] PARALLEL FORCES. 131 

But, in reference to the point B, we have 

F x .ab = F 2 .bc; 

and because ac and A O are cut by parallel lines, we have 

db\lG\\AB\BC\ 

which, by means of the preceding proportion, gives 

ah : ho : : x' — x x : x % —x'\ 

and this, in the above equation, gives 

F 1 (x'-x 1 ) = F i (x 2 -x'); 
or, 

F x x' -FM^F&i-F&T; 

which, by transposing, gives 

W+^x'^Fm + Fm; 
.\R 1 x , = F 1 x l +F^ i . 
In a similar way we would find 

B^'^R^x' + FiXs 

ZZZF^+F^+F^'y 

and, finally, 

Ex = F& + F$s % + etc. = 2Ex>\ 
and similarly, 

Ry = F{y l + F 2 y 2 + etc. = 2Fy. 
From these equations, we have 
-_2Fx 

y - R • 

202. If the System "be referred to Three Rectangu- 
lar Axes, x, y, z 9 in which a, #, and 7 are the angles which 



139 STATICS. [208.J 

the unes of action of the forces make with the respective 
axes ; then the equations for equilibrium will he 

X = cos a$F= R cos a ; 
Y = cos f32F=R cos /?; 
Z = cos ySF= R cos 7 ; 

R = SF; 

Rx = SFx; Ry = SFy; RZ=2Fz; 

The last three equations are the moments of the forces 
in reference to the respective coordinate planes. 

If the given forces are in equilibrium in reference to 
translation, we have 

R = 0; 

and, if they are also in equilibrium in reference to rota- 
tion, we have 

XFx = ; 2Fy = ; SFs ^ 

- o _ _ 

hence, the centre of an equilibrated system is indetermi- 
nate. 

203. Centre of a Mass. — The centre of the mass 
which constitutes a body, is a point so situated that, if its 
distance from, any axis he multiplied by the entire mass, 
the product will equal the sum of the products obtained 
by multiplying each elementary mass by its distance from 
the same axis. 

This point will be determined when its position in 
reference to three rectangular planes is known. 



r SOSJ PAEALLEL FOKCES. 138 

Let M = the total mass of the body ; 
m = an elementary mass ; 
x, y, z, the coordinates of any element, and 
x,y, I, the coordinates of the centre of the mass ; 
then, according to the definition, we have 
Mx = %m%, 
My = Zmy, 
Mz = 2mz. 
These equations are applicable to several bodies If 
the origin of coordinates be at the centre of the mass, we 
have 

aj = 0; p = 0; z = 
.\ 2mx = ; 2my = ; %ms = 0. 

EXAMPLES. 

1 Two parallel forces whose magnitudes are 6 and 11, 
acting in the same direction upon a rigid line, have 
their points of application, A and I>, 5 feet from 
each other ; required the point of application of the 
resultant. 

2. In the preceding example, find the point of application 

of the resultant if the forces act in contrary direc- 
tions. 

3. If the weights 2, 3, 4, and 5 lbs. act perpendicularly to 

a straight line at the respective distances of 2, 3, 4, 
and 5 feet from one extremity, what will be their 
resultant and its point of application ? 

4. Let the weights 3, 4, 5, and 6 act perpendicularly to a 

straight line at the points A, JB, O, and D, the dis- 
tances AB — 3 feet, BO — 4 feet, and AD = 5 feet ; 
required the resultant, and the distance from A to 
the point of application E of the resultant. 



134 STATICS. [20a] 

5. If two parallel forces, P and F, act in contrary direc- 
tions at the points A and B, and make an angle <£ 
with the line AB ; find the moment of each in refer- 
ence to the point of application of the resultant. 

EXERCISES. 

1. Has a statical couple a centre of force ? 

2. Will several parallel forces always be in equilibrium if the sum of 

their moments is zero ? 

3. When may the resultant of parallel forces be zero, and the system not 

be in equilibrium ? 

4. If a system is in equilibrium why may the centre of force be at any 

point ? 

5. If the mass of a body is homogeneous, will the centre of the mass be 

at the geometrical centre of the body ? 

6. If, in a sphere, the density varies directly as the distance from the 

centre in all directions, will the centre of the mass be at the centre 
of the sphere ? 

7. State different laws according to which the density in a sphere may 

vary, and have the centre of the mass at the centre of the sphere. 



CHAPTER IX 

CENTRE OF GRAVITY. 

204. The lines of action of the force of gravity con- 
verge towards the centre of the earth ; but the distance 
of the centre of the earth from the bodies which we have 
occasion to consider, compared with the size of those 
bodies, is so great, that we may consider the lines of action 
of the forces as parallel. The number of the forces of 
gravity acting upon a body may be considered as equal to 
the number of particles composing the body. 

205. The Centre of Gravity of a body m&y be defined 
as the centre of the parallel forces of gravity acting upon 
the body ; and hence the centre of gravity of bodies may 
be found in the same way as the centre of parallel forces. 

206. The Resultant of the Force of Gravity equals 
the weight of the body ; or 

B = W. 

207. If a body be supported at its centre of gravity, and 
the body be turned about that point, it will remain in 
equilibrium in all positions, for it will be equivalent to 
turning the forces through the same angle. 

208. Proposition. — If a body be suspended at any 
point, then, for equilibrium, the vertical through the cen- 
tre of gravity will pass through the point of support. 

Let the body be suspended at the point c, the centre of 
gravity being at a. If the vertical through a does not 
pass through c, join the points c and a by the line ca, draw 



136 



STATICS. 



[209, 210. j 



the vertical ch, and the horizontal ah. The weight W maj 
be represented by the line ch, of which the components 
are ha and ca. The component ha will cause the body to 
turn about c, causing the centre a to approach' the vertical 
ch. If a be in the vertical ch, either above or below the 
support, there will be no horizontal component, and the 
body will be in equilibrium. 






Fig. 78. 

209. Stable Equilibrium. — In stable equilibrium, if 
the body be turned slightly from its position of rest, it 
will tend to return to its former position. Thus, the body 
represented in Fig. 77 is in stable equilibrium. It appears 
also from Figs. 76 and 77, that when the equilibrium is 
stable, and the body is turned about the support, the centre 
of gravity will be raised. In this case, therefore, the cen- 
tre of gravity is the lowest possible. The measure of the 
stability is the amount which the centre of gravity is raised 
in overturning the body. 

210. Unstable Equilibrium. — A body is in a condition 
of unstable equilibrium, if, when it is turned slightly from 
its position of rest, it departs, or tends to depart, farther 
from that position. This is illustrated by Fig. 78. In 
this* case the centre of gravity will fall, when the body is 
turned about its support. 



211-213.] 



CENTRE OF GRAVITY 



137 



211. Indifferent Equilibrium is that condition, in which 
a body will remain in equilibrium after being slightly dis- 
turbed. A sphere or cone resting on a horizontal plane 
is an example of indifferent, or neutral, equilibrium. 

212. Trial Methods. — The preceding principles enable 
one to determine, in an experimental manner, the centre 
of gravity of bodies. Thus, 

in Fig. 79, let the body be 
carefully balanced upon a 
knife-edge, and, when bal- 
anced, the line of support be 
carefully marked upon the 
body. Then balance it in a 
similar way along another 

line ; the intersection of the lines will be vertically under 
the centre of gravity, and, if the body be a thin plate, we 
say that the centre of gravity is at their intersection. 

213. If a body be suspended at a point a, the centre of 
gravity will be vertically under it. Draw the vertical 
line ab upon (or within) the body; then suspend it at 




Fig. 79. 





Pig. 80. 



Fig. 81. 



another point e, Fig. 81, and mark the vertical line cd. 
The centre of gravity will be at g, the intersection of the 
lines ab and cd. 



138 STATICS. [213,S14.J 



EXERCISES. 

1. If a carriage stands upon a side hill, what condition must be fulfilled 

in order that it shall not overturn ? What must be the condition 
that it shall overturn ? 

2. A man stands upon a floor ; how far can he lean forward or backward 

and not fall over ? 

3. When a man moves his head forward, what other motion must his 

body have that he may remain in equilibrium upon his feet ? 

4. Why will not a table be as stable when standing upon two legs as 

upon three ? 

5. Why is it more difficult to overturn a body like Fig. 77 than it is 

one like Fig. 78, the bodies being of equal weight ? 

6. If a book be suspended at one corner, why will its edges be inclined 

to a vertical ? 

7. May a body be in a state of neutral equilibrium in reference to a 

disturbance in one direction, and stable in reference to another ? 





Fig. 82. Fjo. 83. 



8. Explain how the toy shown in Fig. 82 may be in stable equilibrium. 
0. Explain how the toy horse shown in Fig. 83 stands upon the post 
without falling off. 

Centre of Gravity of Heavy Particles. 

214. Centre of Gravity of two Particles. — Let P 

be the weight of a particle at A, and 

O x ® W, that at C. The centre of gravity 

Pxo * ^ will be at some point B y on the line join 

ing A and O. Hence, 



1215,816.] 



OBNTEE OF GBAVITT. 



139 



but, 



P.AB = W.BC\ 



AB + BC=AC; 
which, combined with the preceding equation, gives 

W 



AB = 



Li 
then 



W+P 
P = W, 



AC. 



AB = iAC. 

215. Centre of Gravity of several Heavy Parti- 
cles. — Let w t , w 2 , w 3 , etc., be the weights of the particles. 
Join w x and w z by a straight line and find their centre of 
gravity A, as in the preceding article. Join A with w 2 
and find the centre of gravity B, which will be the centre 
of gravity of the three weights w z , w ly w 2 . In a similar 
way find O, the centre of gravity of the four weights. In 
this way the centre of gravity of any number of weights 




Pio. 85. 




Fig. 86. 



may be found. It is not necessary that the weights be in 
one plane ; they may be distributed in any manner in 
space. 

216. If the Positions of the Particles are referred 
to Three Rectangular Axes, let 



140 STATICS. [317.) 

w l9 w 2i i0 3 , etc., be the weights of the respective 

particles, 
#i> Vi? Zi the coordinates of w t and similarly for vo* 

w z , etc., 
x, y y s the coordinates of the centre of gravity of 

all the weights, and 
W the sum of all the weights ; 
then, we have 

W= to 1 + w 2 + w 3 -f etc. — Sw ; 
and, according to Article 202, we have 

%wx _ Hwy _ 2wz 

EXAMPLES. 

1. Two particles are joined by a straight line ; if one is n 

times as heavy as the other, find the position of the 
centre of gravity. 

2. If three equal particles are at the vertices of a triangle, 

find the position of the centre of gravity. 

3. If the weights of three particles are as 1 to 2 to 3, and 

are placed at the vertices of an equilateral triangle, 
find the position of the centre of gravity. 

4. If four equal weights are at the vertices of a triangular 

pyramid, find the position of the centre of gravity. 

Centre of Gravity of Lines, 

217. Straight Lines. — By a line, we here mean a mate- 
rial line, whose transverse section is very small, such as a 
very fine wire. 

The centre of gravity of a uniform straight line is at 
its middle point. 



[818.] 



CENTRE OF GRAVITY. 



141 




For, we may conceive it to be composed of pairs of 
particles, each of which is at the same distance from the 
middle point ; and as this point will be the common centre 
of gravity of all the pairs, it will be the centre of gravity 
of the line. 

218. Centre of Gravity of the Perimeter of a Tri- 
angle. — Let AB G be the triangle. The centre of gravity 
of the sides will be at 
their middle points, D, 
E, F. Join these points. 
The weight at E will be 
to that at D as the length 
of GB is to the length 
of A G. Divide the line 
BE at the point G, so 
that 

BG:GE::BG:AG, 

then the point G will be the centre of gravity of the two 
lines AG and GB. Similarly, I will be the centre of 
gravity of B G and AB, and H, that of AB and A G. The 
construction gives 

BF=iBG; BE=%AB; FE=\AG; 
hence, the preceding proportion gives 

DG\GE\\\BG \\AG 
: : DF : FK 

Therefore, the line drawn from F to G will bisect the 
angle F; and, similarly, for DI and EH. 

The centre of gravity of the three sides of the triangle 
ABG, will be in the line GF; and, similarly, it will be in 
the lines Z>/and HE; hence, it will be at their intersec- 
tion, which will be the centre of the circle inscribed in the 
triangle BEF, 



142 STATICS. 1219,220.] 

219. Symmetrical Lines. — The centre of gravity y>i 
lines which are symmetrical in reference to a point, will 
be at that point. Thus : — 

The centre of gravity of the circumference of a circle, 

or an ellipse, is at the geometrical centres of those figures : 

The centre of gravity of the perimeter of an equilateral 

triangle, or of a regular polygon, is at the centre of the 

inscribed circle : 

The centre of gravity of the perimeter of a square, rec- 
tangle or parallelogram, is at the intersection of the diag- 
onals of those figures. 

220. Centre of Gravity of a 
Circular Arc— Let ABO be an 
arc of a circle, its centre, B its 
middle point, and AC its chord. 
The centre of gravity will be on 
js^ the radius BO, at some point c, 

FioTk such that (see Article 234), 

Arc AB O : radius BO:: chord A O : Oc. 
BO. AG 




Oc = 



ABO 



EXAMPLES. 

1. Find the position of the centre of gravity of the edges 

of a rectangular box. 

2. Find the position of the centre of gravity of the edges 

of a regular pyramid having a square base, and 
whose altitude equals the length of one side of the 
base. 

3. Find the centre ot gravity of the semi-circumference 

of a circle, % r 

Ana, 0c = — . 



[221,222.] CENTRE OF GRAVITY 143 

4. In Fig. 88, if the angle AOC= 60°, find the position 

of the centre of gravity of the arc ABC. 

Am. Oc=. — . 

5. Find the distance from the centre of a circle to the 

centre of gravity of a quarter of the circumference 
of the circle. 

Arts. Oc = 2V2-- 

IT 

Centre of Gravity of Surfaces. 

221. Definition. — A surface here means a very thin 
plate or shell. 

222. The Centre of Gravity of a Plane Triangle 

is in the line joining the vertex with the middle point of 
the base, and at one-third the length of the line from the 
base. 

Let ABC be a triangle. Consider the triangle as com- 
posed of an indefinitely large number of straight lines 
parallel to the base AB. The centre 
of gravity of each of these lines is at 
its middle point ; hence, the centre of 
all of them will be at some point in 
the line passing through their centres; 
which point will be the centre of grav- 
ity of the triangle. Let D be the mid- 
dle point of AB ; join C and J), then 
will the centre of gravity of the tri- 
angle be in the line CD. Similarly, it 
will be on the line AE, drawn from A 
to the middle point of BC; hence, it will be at g, the inter- 
section of DC and AE. To find the distance Cg, draw 
DE, then will the similar triangles DEg and AgC givv 




144: 



STATICS. 



[333, 224. 



= i 



DE _gD 

AG~ gC 

.\gC=2gD. 

Adding gJD to both members, gives 

gC+gZ> = DC=3gD 

.:gD = iDC. 

Similarly, 

gE=\AE. 

We may readily find that, the perpendicular distance of 
the centre of gravity from the base, equals one-third of 
the altitude. 

223. Symmetrical Figures. — The centre of gravity of 
the surface of a circle, or of an ellipse, is at the geomet- 
rical centre of the figure ; of an equilateral triangle, or a 
regular polygon, it is at the centre of the inscribed circle ; 
of a parallelogram, at the intersection of the diagonals ; 
of the surface of a sphere, or an ellipsoid of revolution, at 
the geometrical centre ot the body ; of the convex sur- 
face of a right cylinder, at the middle point of the axis 
of the cylinder. The centre of gravity of the convex sur- 
face of a regular right pyramid, 
or of a right cone having a circle 
for its base, is on the axis of the 
figure at one-third the altitude 
from the base ; for, the surface 
may be considered as composed c f 
triangles having a common apex. 
224. To find the Centre of 
Gravity of a Part of a Body, 
when the centre of gravity of the whole and of the remain- 
mgpart are known. 

Let AB be one area, CD the other, O the centre of 




Fig. 90. 



[225,226.] OBNTKE OF GRAVITY. H5 

gravity of the former, o that of the latter, and 6 the centre 
of gravity of the part remaining after removing the area 
CD. The point c will be on the line through oO. Take 
A as the origin of moments. The weights are directly 
proportional to the areas, and the moment of the whole 
equals the sum of the moments of all the parts, hence 

Area AB x A 0=Area CD xAo + (Area AB—Area CD) 

xAc 
Area AB xAO — Area CD x Ao 



.\ Ac = 



Area AB—Area CD 



The same formula will apply to lines and volumes by 
simply substituting line or volume for area. 

225. Irregular Figures. — Any figure may be divided 
into rectangles and triangles, and, the centre of gravity of 
each being found, the centre of gravity of the whole may 
be determined by treating it as if it were an aggregation of 
particles, as in Articles 215 and 216. 

226. The Centre of Gravity of a Zone is at the mid- 
dle jpoint of the line joining the centres of the upper and 
lower bases of the zone. 

It is proved in Geometry that, on the same or equal 
spheres, zones are to each other as their altitudes ; hence, 
if the zone be divided into an indefinite number of paral- 
lel zones, and all be reduced, or contracted, to the axis of 
the zone, it will form a line of uniform weight ; and 
hence, the centre of gravity will be at the middle point of 
the line. 

EXAMPLES. 

1. If a line be drawn parallel to the base of a triangle, 
dividing it into equal areas, will it pass through the 
centre of gravity of the triangle \ 
7 



14:6 



STATICS. 



[226.] 



2. If a line bisects the vertical angle of a triangle, in what 

cases will it pass through the centre of gravity, and 
in what cases will it not ? 

3. If the bases of two triangles are in the same line, and 

their vertices are in a line parallel to the bases, show 
that the line joining their centres of gravity will also 
be parallel to the bases. 

4. In Fig. 90, if the circles are tangent to each other in 

ternally at A, find the distance from A to the centre 
of gravity c, after the smaller circle has been re- 



moved. Let B = AO: r = Go. 



Ans. Ac 



B 2 -r>' 



5. Find the centre of gravity of the remainder of a square 

after one- quarter of it has been removed from one 

corner. 

5 
Am. AB = qAG. 



\ 

c 






k \ 

\ 



¥ y 



Fig. 91. 



6. Find the centre of gravity of a trapezoid. 

It will be on the line joining the centres of the two 



A w rJ7n AB+%DE 



|227.] 



CENTRE OF GRAVITY. 



147 



Find the centre of gravity of the surface of a right 
cone having a circular base, including the base. 
Let r be the radius of the base and h the altitude ; 
find the distance from the apex. 

Ans. — . — h. 

Vr*+h 2 +r 



Centre of Gravity of Volumes. 

22.1. Triangular Pyramid. — The centre of gravity of 
any triangular pyramid is on the line joining any 
apex with the centre of gravity of the opposite face, and at 
a point three-fourths the length of the line from the apex. 

Let A-BGD be a 
triangular pyramid. 
Suppose that it is di- 
vided into infinitely 
thin slices, bed, parallel 
tothebase,^6 r Z>. The 
centre of gravity of the 
pyramid will be on the 
line passing through 
the centres of all the 
slices. Let F be the 
centre of gravity of the 
base, then will the cen- fw. se. 

tre of gravity of the pyramid be on the line AF. Simi- 
larly, it will be on the line BG drawn from the apex B to 
the centre of gravity of the opposite face, and hence, at 
their intersection R. Join i^and G, and the similar tri< 
angles FGE and AEB give 

EG_ FG 
AE~~ AB' 




148 



STATICS. 



[228, 229. J 



But, EG = \AE\ 

.-. FG = iAB. 
The similar triangles, FGH and AHB % give 
FG FH 
AB ~ AH' 
in whichj substitute the value of FG, and it gives 

BFH=AH. 
Add FH= FH, 

and we have 4:FH= AF; 

.\ FH= \AF. 
228. The Centre of Gravity of any Pyramid or 
Cone is on the line joining the apex with the centre of 
gravity of the base and at one-fourth the distance from 
the base. 

That it will be on this line is evident from the preced- 
ing Article. The pyramid may be divided into triangular 
pyramids, and the centre of each will be in a plane paral- 
lel to the base and at one-quarter the altitude from the 
base ; hence, it will be at the point where this line inter- 
Beets the plane. The position for the cone is found in the 

same way, for, the cone may be 
considered as composed of an in- 
definite number of pyramids. 

229. Problem. — Find the 
centre of gravity of a spherical 
sector generated by the revolu- 
tion of the circular sector AGO 
about the axis G G. 

It will be on the axis G G. If 
we consider that the spherical 
gee tor is composed of an indefinite number of cones, having 
their bases in the surface of the sphere, and their apices at 




[230.] CENTRE OF GRAVITY. 149 

the centre G of the sphere, the locus of the centre of gravity 
cf all the cones will be in a spherical surface EEE, having 
G for a centre and radius GD = f of GA. The centre of 
gravity g of this surface will be the centre of gravity of 
the spherical sector ; but, according to Article 226, Eg ia 
one-half of the altitude EX of the spherical surface 
DEE. But, from the figure, we have 

EE=$GH; 

.:Eg = %GH; 
and 

Gg= GE+Eg 

=iGG+%GH 
= i(GG+§GB). 
But, 

Gg = GG-Gg 

= GG-IGG-%GE 
= %(2GG- GH). 

230. Problem. — To find the centre of gravity of a seg- 
ment of a sphere. 

Let A GE, Fig. 94, be the segment of a sphere, and a 
point g' on the line GH be its centre of gravity. Taking 
the origin of moments at the centre G, we have 

Vol. ofSeg. x Gg' = Vol. of sector AGE Gx Gg 
- Vol. of cone AEG xiGH; 
or, 

Tr{GH)\GG-%GH).Gg' = ZTr(GG)\^.WC>Cg 
—iriAEf.iGEr.iGK; 

. r , _ 8( GG)\ GH. Gg - S(AHf. GH* 
•'• Vg ~ 12(GHf (GG-iGH) 



150 STATICS. |231, 232.| 

EXAMPLES. 
1 ; Find the centre of gravity of a hemisphere. 

It will be on the radius perpendicular to the base 
of the hemisphere, and, according to Article 229. 
at a distance f the radius from the centre. 

2. Find the centre of gravity of the remainder of a sphere 

whose radius is B,, after another sphere, whose radius 
is r, is taken from it, the two spheres having a com- 
mon tangent plane. 

3. Find the distance from the centre of a sphere to the 

centre of gravity of a segment of the sphere of one 
base, the chord of the segment being J the radius. 

4. A cone is suspended at a point in the circumference of 

the base ; required the inclination 6 of the axis to 
the horizontal. Let the radius of the base be 2 
inches and the altitude 8 inches. 

5. In the preceding example, what will be the relation 

between the radius of the base and altitude of the 
pyramid, if the axis is inclined 30° % 

Centroharic Method. 

231. The two following theorems are by some accredited 
to Guldinus and by others to Pappus, one or the other of 
whom is supposed to have discovered them. 

232. Theorem I. — The surface, generated by the revo- 
lution of aline about an axis fixed in the plane of the line, 
is equivalent to the product of the length of the line into 
the circumference passed over by the centre of gravity of 
the line. 

Let AN be a plane curve, and YY the fixed axis in 
the plane of the curve. Draw any number of equal 




[232.] CENTROBARIC METHOD. 151 

chords, AB,BC\ etc., and from their middle points, a u a^ 
etc., draw the perpendiculars aj)^ aj) 2 , etc., to the axis 
77. The revolution of 
the curve about the axis will 
generate a double curved 
surface, and the polygon, 
several frustra of cones in- 
scribed within the former 
surface. 

The surface generated by 
AB will be (see Geometry) 

2irOibi X AB. 

Similarly, the entire sur- 
face generated by the polygon will be 

27r(& 1 5 1 x AB + aj>2 x B 0+ etc.). 

Let g be the centre of gravity of all the lines AB, BO, 
etc., and go, a perpendicular to 77, then will the mo- 
ments of the lines in reference to the axis 77 be 
gc{AB + BC+ etc.) = AB. aA + BO. a 2 h 2 + etc. 

Multiplying both members of this equation by 27r, gives 

27rgc(AB + BC+ etc.) = 2ir(AB.a 1 b 1 + £C.a£ 9 + etc.), 
the second member of which is the surface generated; 
hence, 

2irgG -^perimeter of the polygon = surface generated, 
in which 2>irgc is the circumference described by the cen- 
tre of gravity of the perimeter. 

Inscribe in the curve another polygon of double the 
number of 6ides, and so on indefinitely ; the limit cf the 
polygons is the arc, and the limit of the surface is the 
double curved surface ; but, the preceding equation is 
true for any number of sides, and hence, will be true of 



152 



STATICS. 



[ 233. 1 



the limit. Let g v be the centre of gravity of the arc, and 
g x c x the perpendicular upon the axis ; then the equation 
becomes 

^Trp'itfi X length of arc = surface generated by the arc. 
The theorem is evidently true for a single line, or foi 
several lines of unequal length. Q. E. D. 

233. Theorem II. — The volume, generated by the revo- 
lution of a plane area about a fixed axis in its plane, is 
equivalent to a prism whose base is 
the area revolved, and altitude, the 
length of the circumference passed 
over by the centre of gravity of the 
area. The plane area must lie wholly 
on one side of the axis. 

Let LAB . . . FQ be a plane area, 
YY the fixed axis in the plane of 
the area. Divide LQ into equal 




Fig. 96. 



parts, LM, MN, etc., and draw the 
ordinates LA, MB, etc., perpendicu- 
lar to YY, and from A, B, etc., 
draw parallels to YY, forming rectangles, as shown in 
the figure. When the figure revolves the rectangles will 
generate cylinders, and the curve, a double curved surface. 
The volume of the cylinders will be 

tt^IZ 2 . LM+irBM\ J/iF+etc. 
Let g be the centre of gravity of all the rectangles, and 
the ordinate to the centre of gravity of each being equal 
to one-half the length of the side of the rectangle, we 
have for the moments of the rectangles in reference to the 
axis YY, 

gc x area of all the rectangles = AL . LM . \AL-^ 
BM.MJT.iBM+etc. 



[234.] APPLICATIONS. 153 

* Sirgcxarea of all the rectangles = ttAL 2 . LM-V 
wBM\MN '+ etc. 

= Vol. of all the cylinders. 

If now the divisions in LQ be increased indefinitely, 
the limit of the rectangles will be the area of the curve, and 
the limit of the cylinders will be the volume of revolution. 
Let g x be the centre of the plane area, then we have 

Vol. of revolution = ^irg-^ . area of the plane 

curve, 
— the area of the curve x by 
the distance described by 
the centre of gravity of 
the area. Q. E. D. 



Applications. 

234. Problem. — Tojmd the centre of gravity of a cir- 
cular arc. 

Let ABC be a circular arc whose centre is O; the 
centre of gravity will be at some point c on the radius OB 
drawn to the middle point, B, of the arc. 
Through O draw the axis yT~parallel to the 
chord AG, and conceive the curve to be 
revolved about this axis, generating a zone. 
The area of the zone will be (see Geometry) 

2ttOB.AC. 
According to Theorem 1, we have 

Arc ABC. 2ttOc = Area of the zone = 2ttOB.AC; 
~ UG ~ ABC * 




154: 



STATICS. 



[235.] 



235. Problem. — Find the distance from the centre of a 
circle to the centre of gravity of a sector of the circle. 

Let AGBG be the sector generated by the rotation of 
the line A G about the centre G. Let 6 be the angle A GG, 

and r the radius AG; then will 
the arc AGB — 2r0. Accord- 
ing to Theorem 1 we find that 
the area of the sector will be r^Q. 
If the sector be divided into 
an indefinitely large number of 
sectors, each may be considered 
as a triangle whose centre of 
gravity is at two- thirds of its alti- 
tude from the centre G. With 
\r, describe the arc DEF r ; this 
arc will be the locus of the centre of gravity of all the 
small sectors ; and the centre of gravity of all of them, or 
the sector A GB will be at g, the centre of gravity of the 
arc DEF. According to Article 234: we have 




a radius D G, equal to 



Og 



2 x %r sin 6 . f r _ 2 r sin 

— "8 



2x%r6 



e 



EXAMPLES. 



1. If AGBG is a semi-circle, prove that Gg is -J - 



IT 



2. If g', Fig. 98, be the centre of gravity of the circular 

segment AGB, find the distance Gg' . 

3. Find the volume of a sphere. 

4. Find the volume generated by the- revolution of the 

circular segment A GB about an axis through G and 
parallel to the chord AB. 

5. Find the volume generated by the circular sector GAGB S 

about an axis through G and parallel to AB 



CHAPTEK X. 

SOLUTION OF PROBLEMS ACCORDING TO THE PRINCIPLES OF 
ENERGY. 

Problems in which the Solutions depend upon 
Potential Energy. 

236. Energy represented by the Three States of 
Equilibrium. — According to Article 210, when a body ie 
in the condition of unstable equilibrium, the centre of 
gravity is in the highest position. In this condition its 
potential energy is a maximum, that is, it is in a condition 
to do the most work. When the centre of gravity is 
lowest, the body is in a condition of stable equilibrium 
(Article 209), and its potential energy is a minimum, that 
is, it is in a condition to do the least work. In neutral 
equilibrium, the potential energy, for successive positions 
of the body, remains constant. 

237. To find a curve such that a heavy bar AE, rest- 
ing against it and against a vertical 

plane DE, will be in equilibrium in 
all positions, there being no friction 
on the surfaces. / 

F / 

Let g be the centre of gravity of c r | — 7y 

the bar. If the bar is prismatic and V \ / 

homogeneous, g will be at the middle jj^ 

of the length, but, in other cases, it _ 99 B 

may be at any other point along the bar. 

This is a case of indifferent equilibrium, and hence, the 
centre pf gravity is neither raised nor lowered by a change 



156 



SOLUTION OF PROBLEMS. 



[238.] 



of position ; in other words, its locus will be in a horizon- 
tal line. Therefore, if the centre of gravity g be moved 
along the line CD, and the end E be kept constantly 
against the vertical DE, the end A will trace the curve. 
But this is equivalent to the well-known method of con- 
structing an ellipse by means of a trammel. Hence, the 
curve GAD is an ellipse. 



[Let 



then, 



and 



or, 



AB=l&ndAg = ^l; 
z = BF,y = FA; 

Fg = fr, 

FA* + Fg*=Ag*; 



which is the equation of the ellipse, 
will be found by letting Ag — n. Eg.] 



A more general solution 



238. Required the form of a curve such that a heavy 
bar resting against it and against a smooth jpin above the 
curve, will be in equilibrium in all positions. 

Let AD be the bar, D the 
position of the pin, and ABO 
the required curve. 

This is also a case in which 
the potential energy is con- 
stant j hence, the centre of 
gravity will be found in a 
horizontal line gg\ passing 
through the centre of grav- 
ity, g, of the bar in the vertical position. The curve may 
therefore be constructed by drawing any number of radial 
lines DD, DA, etc., through D, intersecting them by the 
horizontal line gg\ and laying off on the radial lines below 




[239, 240.] POTENTIAL ENE&GrY. 157 

the horizontal the constant distance gB = g'A, etc. The 

curve is called the conchoid of Nicomedes. 

[Let 

Ag' = Bg = a; gD = c; ADB=d\ AD = p; 
then, 

AD = Ag'+g'D 

= Ag' + Dg sec 6 ; 
o = a + c sec 6 ; 
which is the polar equation of the curve.] 

239. A cord of given length is suspended at two points 
in the same horizontal ; required the form of the curve 
when the centre of gravity is the lowest. 

The cord, being perfectly flexible, will naturally assume 
the position of stable equilibrium, 
and its potential energy will be a 
minimum ; that is, its centre of grav- 
ity will be the lowest possible. The 
curve assumed by such a cord is called a Catenary. If the 
cord be of variable density, it will still assume the position 
in which the centre of gravity is lowest. 

[The equation of the catenary is found by higher mathematics. 

If w = the weight per unit of length of the cord ; t = the ten- 
sion at the lowest point of the cord ; e = the base of the Naperian 
system of logarithms ; x horizontal and y vertical ; the origin of 
coordinates being taken at the lowest point, then 




[ wx _!£^~l ' 



240. A curve of given length is revolved about the line 
passing through its extremities ; required the form of 
the curve such that the surface generated shall be a 
maximum. 



158 



SOLUTION OF PROBLEMS. 



[341.] 



The area will equal the length of the curve multiplied 
by the distance passed over by the centre of gravity of the 
curve. The curve being of constant length, the area 
will therefore be greatest when the distance of the centre 
of gravity of the curve from the axis of revolution is 
greatest ; hence the curve must be a catenary. 

241. Two cylinders of unequal radii, but of the same 
material and length, are placed in a larger hollow cylinder; 
find the^ position when in equilibrium. 

Their position will be that in which the potential energy 
is least ; hence, their centre of gravity will be lowest. 

Let g be the centre of 
gravity of the cylinders when 
in contact. If they be rolled 
in the hollow cylinder, re- 
maining in contact with each 
other, the centre of gravity 
will describe a circle MgN t 
about the centre C of the 
hollow cylinder, and when 
they are in the position of equilibrium, the point g will 
be vertically under the centre C. 




Fig. 102. 



To find the angle ECg, let R - AC, r 2 = FB, 
r x = AK 

Then, EF= r x + r 2 = the sum of the radii of the two 
cylinders ; and the equation of moments gives 

W l .Eg= W,.gF; 



and from the figure we have 



1242.] 



and, 



POTENTIAL ENERGY. 

Eg + gF= r x + r 2 
CE=B-r-r t . 



159 



242. Required the effort necessary to maintain a oody 
on a smooth, inclined plane, the effort being exe?*ted par- 
allel to the plane. 

Let the effort be exerted by a body acting vertically, as 
in Fig. 103 ; then will the centre of gravity of the two 
bodies be in the same horizontal line for all positions of 
the body on the plane. 

Let A C be the inclined plane, on which the weight P 
is held in position by the weight W, all without friction. 
Let a and d be the centres of the 
bodies respectively in one posi- 
tion, then will their centre of 
gravity be at some point g, on 
the line joining their centres. 
If the bodies are moved into an- 
other position, having their cen- 
tres respectively at b and e, their 
centre of gravity will be in a 
horizontal line through g. 

Since the energy is constant, the potential energy gained 
by raising the weight P will equal that lost by lowering 
W, or, in other words, the product of P into the vertical 
distance through which it has been raised, equals W into 
the vertical distance through which it has been lowered. 




Fiq. 103. 



160 SOLUTION OF PEOBLEMS. [243, 244. j 

Drawing the horizontal line be, and the vertical line ao i 
we have 

P. ac = W.de; 

but, 

ac = db sin cba 

= db sin A 
= de sin J. 
, BG 
- ae AO ; 
which, substituted in the preceding equation, gives 

P.BG = W.AC; 
or, 

P:W::AO: BO; 

that is, the effort is to the resistance as the height of the 
•plane is to its length. 

243. Determine the conditions of equilibrium of a 
single pulley. 

In one position let the weight W be at d, and P, at a ; 
and in another position, P at b y and W at c. 
/^"N The centre of gravity will be at the same point 
^y g, in both positions ; hence, we have 

c P.ab = W.dc; 

* but, 
P W ab = dc; 

FI °' 104 - .:P=W; 

that is, £Ad effort equals the resistance when there is no 
friction. 

244. Determine the conditions of equilibrium of the 
straight lever. 

Let AB be a bar in a horizontal position, having weights 
P and W suspended at its extremities ; it is required to 
find the point C\ upon which they will balance. Their 



[244.] 



POTENTIAL ENERGY. 



161 




Fig. 105. 



centre of gravity, g, will be in the line joining the centres 

of the bodies, and vertically under the required point G. 

Let the arm be turned 

into the position ab, 

the perpendiculars da 

and eb from a and b 

be dropped upon 

AB ; then will the 

weight W have been 

raised a height equal 

to he, and P will have 

fallen a distance ad. 

Since there is equili- 

rium, we have 

P.da=W.eb. 

The similar triangles ad G and beG, give 

da _ aG _AG 
~eb~ b~G~BG ; 

and, by combining these equations, we find 
P.AG=W.BG; 

P:W::BG:AG; 

that is, the effort is to the resistance inversely as the arm 
of the effort is to the arm of the resistance. 

In this problem the centre of' gravity of the bodies re- 
mains at g for all inclinations of the arm AB ; for the 
similar triangles adG and Geb, give 

dG: Ge::aG: Gb; 
but dG and Ge are the arms of the forces in the new posi- 
tion ; hence, they are proportional to the original arms. 
When the support is not in the line of the points of attach- 
ment of the weights, the centre of gravity will change its 



162 



SOLUTION OF PROBLEMS 



1245.J 




W\ 



Fig. 106. 



position as the arm is rotated, as will be seen in the fol- 
lowing problem : 

245. Tofirtd the conditions of equilibrium of the bent 
lever. 

Let A G and BGbe the arms of the lever, the support, 
or fulcrum, being at G, and the weights suspended as 

shown in the figure. 
,d The centre of grav- 
ity of the weights, 
when in equilibri- 
um, will be in the 
line joining the cen- 
tres of P and W, 
and at a point g y 
vertically under the 
point of support G. 
Let the lever be 
turned through a small angle, then will the end A de- 
scribe the arc A A', and B, the arc BB\ and the weights 
will be found in the positions P' and W. The centre 
of gravity will be raised to a point g' ; hence, when the 
bodies are left to themselves they will return to their 
former position. 

We may, however, determine a practical formula fo T .his 
case by assuming that they remain in equilibrium when 
the lever is turned through an exceedingly small angle. 
For this case, we consider the arcs A A' and BB' as 
straight lines. Draw the horizontal lines Ala and B'b, 
then will the potential energy of P be increased ty an 
amount equal to 

P.Aa; 

and that of TPwill be diminished by an amount equal to 

W.Bb; 



f246.] KINETIC ENERGY. 163 

hence, according to the hypotheses, we have 

P.Aa=W.Bb. 

Through G draw the horizontal CD, meeting the verticals 
through A and B at the points C and D. From the simi- 
lar right angled triangles A A! a and A CG, we have 

Aa __ GO 
AA' " GA' 

Similarly, the triangles BB'b and BGD give 

Eb _ GD 

BB' " £.£ ' 

But the arcs BB' and ^Ll' are proportional to the radii 
GB and GA ; hence, 

.AJ/ _ BB f 

GA ~~ £.# ; 

and, by combining these three equations so as to eliminate 
A a and Bh, we find 

P.GC=W.GD; 

P:W::GD:GC; 

that is, the weights are inversely as their arms / a result 
which agrees with the preceding problem. 

Problems involving Kinetic Energy. 

246. A body falls freely through a height h ; what will 
be the kinetic energy stored in it f 

A body whose weight is W, at a height h above a given 
point, has a potential energy of 

Wh; 

and when it has fallen through this height its energy will 



164 SOLUTION OF PROBLEMS. [247.] 

he changed to kinetic energy. Substituting for h its value 
in terms of v (see Eq. (3), Art. 72), gives 

as given in Article 112. 

If the body had fallen through a portion of the height 
h x , leaving a height h^, through which it may afterwards 
fall, we have the kinetic energy 

K=Wk 1 = iMv i *, 
and the potential energy 

II = Wh 2 ; 
hence, the total energy will be 

K+n=W(h 1 +h 1i ) 
= Wh, 
which is constant for that height. 

247. Two bodies of unequal weights are placed on two 
unequally inclined planes, and connected by a fine inex- 
tensible cord, which passes over a pulley so placed above 
the angle of the planes that the cord will be parallel to the 
planes ; required the equations for their motion, there 
being no frictional resistances, nor resistance of the air. 

Let J. (7 and BG be the inclined planes, W and P the 
positions of the bodies 

when motion begins. Af- SQ\ 

ter a time t let them be ^^^^wS/k 

in the positions represent- w^ jjBp^^f""'" W 

ed by F and W ; the dis- »^ h' r ^lllk[ ' 

tances over which they ^ Jz_ JjJP^ 

have moved being FiG m B 

dc = ab. 

In the initia. position, the centre of gravity of the bodies 



(247.1 KINETIC ENERGY. 165 

will be at some point g, in the line joining their respective 
centres of gravity ; and after a time t it will be at some 
lower position g'. Through g and g' draw horizontal lines ; 
the vertical distance ge between them, will be the height 
through which the common centre of gravity of both 
bodies will have fallen. 
Let 

ge = h; 

then, the potential energy lost will be 

{P + W)h, 

which is a gain of kinetic energy equal to 

Let 

s = db = dc ; 

then will the vertical height through which P has fallen be 

s sin B ; 

and the height through which W has been raised will 

be 

s sin A ; 

and the total potential energy lost will be 

(P8mB-WsmA)s; 
hence, we have 

(P + W)^- = (PemB - W8inA)s; 

from which we find 

o P sin B — TFsin A a ... 



166 



SOLUTION OF PROBLEMS. 



[ 248, 249. 1 



Id this problem the acceleration will be constant ; hence, 
according to Article 24, we have 

s = \ot ; 
which, substituted in the preceding and reduced, gives 



v = 



Psm£ - W&inA 



gt. 



(2) 



P + W 

Eliminating v from the two preceding equations, gives 

P + W 2s" 



= \/[pIi 



sin B — IF sin A ' g 



] 



(3) 



248. If one of the todies as P, in the preceding prob* 
lem, moves vertically, while the other moves on the plane, 

required the formulas for 
their motion. 

This problem may be solved 
in the same way as the pre- 
ceding one; but the results 
may be obtained directly from 
iJ&Jap' the preceding formulas by 
making ^ = 90°. The re- 
sults are 




Fig. 108. 



-A 



P 



p _ Wsm A 
P + W 

IF sin A 



2gsj 



and, 



P + W 
P + 



fft> 



I r p + w 2£*i 

V \_P — WsmA' g\' 



249. If in the preceding problems, both bodies move 
vertically, required the formulas for the motion. 




f250.] KINETIC ENERGY. 167 

Making A = 90°, in the preceding formulas, we find 
the formulas of Problem 3, Article 90. 

250. A vessel is filled with a liquid ; required the veto- 
city with which it will discharge itself through an orifice 
near the oase. 

Let ABE be the vessel, F the position of the orifice. 
Suppose that a small portion, equal to the horizontal slice 
ACB, has been discharged. The 
centre of gravity of the mass will 
have been lowered from some point 
g to another point g\ and the poten- 
tial energy lost is equal to the weight 
of the liquid above the orifice, mul- 
tiplied by the distance gg' . But a 
more simple way of considering 
the problem, is to assume that the centre of gravity of the 
part below the slice ACB remains at g\ and hence, that 
the change in the position of the centre of gravity has 
been produced by the transference of the slice A OB to the 
level of the orifice F Assuming that there are no fric- 
tional resistances, nor resistance from the air, then the 
whole energy will be expended in producing the motion 
of the liquid. 
Let 

S = the horizontal section of the vessel at AB, 

h = the section of the orifice at F, 

x = the thickness of the thin slice AB, 

h = DC = the vertical height of AB above the 

orifice, 
t = the time of the discharge of a quantity of the 

liquid equal to that in the slice AB, 
w= the weight of a unit of volume — say one cubic 
inch of the liquid, and 



168 SOLUTION OF PROBLEMS. [250.J 

v = the velocity of the discharge. 
Then, 

Sx = the volume of the slice, 
wSx = the weight of the slice, and 
{wSx)h = the work accumulated in the slice when it has 

fallen through the height h. 
The quantity which will flow through the orifice in a 

time t will be 

kvt y 

the weight of which will be 



the mass of which is 

wkvt 



and of which the kinetic energy, due to the flow, will be 
hence, 



2 9 



if 
But the weight wSx = wkvt, and by cancelling and re 

ducing we have 

v* = 2gh; 

which is the same as that of a particle falling freely 
through a height h. 

This result is modified in practice on account of the re- 
sistance of the air, friction, and viscosity of the liquid. If 
there is a pressure upon the top of the vessel, we ascertain 
what height of the liquid will produce the same pressure 
per square inch of the upper surface, and add the height 
to the value of A, given in the problem. The height 
which induces the flow is called a head. 



[250.] KINETIC ENERGY. 169 

EXAMPLES. 

1. If, in Fig. 107, P = 25 lbs., W=30 lbs., the angle 

B = 60°, and A = 30°, determine which will move 
down the plane, P or W, and how far they will 
move in 5 seconds. 

2. Determine the relations between the weights P and W, 

and the angles A and B for equilibrium in Fig. 107. 

3. If the angle A = 30° and B — 45°, find the relation 

between P and W so that the acceleration of the 
bodies will be -J- that of a body falling freely. 

4. In Fig. 102, if the radius GA is 3 feet, AE 1 foot, and 

FB 6 inches, and the internal cylinders of the 
same material, what will be the angle ECg for 
equilibrium ? 

5. Solve the problem in Article 248, when there is a con- 

stant friction al resistance on the plane equal to 
fi TFcos A. 

6. In the problem of Article 241, find the angle ECg, 

where the cylinders E and F have the same diame- 
ter, but W x = 2 W 2 . 

7. A vessel is filled with a liquid and kept constantly full ; 

required the time necessary for the discharge of a 
quantity q from an orifice whose section is k, the 
distance of the orifice below the surface of the liquid 
being A feet. 

8 



CHAPTEK XL 

CONSTRAINED EQUILIBRIUM. 

251. A body is said to be constrained when it is pre- 
vented from moving in a particular direction on account 
of a point or line of the body being fixed, or on account 
of the interposition of a body which is considered as im- 
movable. By the term fixed is not meant that the resist- 
ance offered by a point, line, or surface, cannot be over- 
come by any force, but that it will not be equaled by any 
force which may be involved in the problem. The term 
immovable is also considered in the same restricted sense. 
When only a point of the body is fixed, the body will be 
free to rotate about it in all directions ; and when a line 
is fixed the body may rotate about it or slide along it. 

252. Normal Resultant. — If a body be in equilibrium 
on a smooth surface, the resultant of all the forces which 

act upon it must be in the direction 
of the normal to the surface and act 
towards the surface. For, if it be not 
normal, it may be resolved into two 
components, one of which will be 
normal and the other tangential, the 
latter of which would produce mo- 
tion. 

253. Equilibrium of a Body on a Smooth Inclined 
Plane. — Let A G be the inclined plane, o the centre of the 
body, i^the resultant of all the forces which act upon the 
body, and W the weight of the body, Resolve the weight 




[254. J 



CONSTRAINED EQUILIBRIUM. 



171 




Fig. 111. 



W into two components, one, cb, parallel to the plane, and 
the other, ob, normal to it ; and, similarly, resolve the* 
force F into the normal com- 
ponent de, and the component 
oe parallel to the plane. The 
excess of ob over de need 
not be considered, since the 
plane must resist it ; but, that 
there shall not be movement 
along the plane, the compo- 
nent cb must equal oe. 
From the figure we have 

cb = oe sin cob 
= IF sin J. ; 
and 

oe = od cos doe 

= i^COS (j), 

where <j> = doe, the angle between the action-line of the 
force and the plane, hence 

i^cos <f> = TTsin A ; 
W 



cos<£ 



F 



sin^l. 



254. Equilibrium on a Rough Inclined Plane. — 

Let the notation be as in the preceding article. The 
amount of friction will be, according to Articles 107 and 
108, the normal pressure multiplied by the coefficient of 
friction, or, 

fi(ob — de) 
= fi( TFcos A — Fsm cj>). 

If the body is in a state bordering on motion down the 
plane, the upward pull of F along the plane, and the 
friction, will equal the downward pull of W; oj, 



172 



STATICS 



[255.] 



i^cos </>+fi{ Wcos, A— F&in <£) = JPsiu A ; 

from which we find 

7^_ sin A— fj, cos A ^ 
COS0 — yLtsin^> 

If the value of F is such that the body is in a state 
bordering on motion up the plane, the component of F, 
parallel to the plane, will equal the component of W, also 
parallel to the plane, plus the friction ; or 

i'cos </> = IF sin A +fi( Wcos, A— F&m <j>) 

Tr _ sin J.-J-yucos A w 
cos <j> + fj, sin <j) 

255. Determine the conditions of equilibrium of a 
homogeneous disc, having a hole cut in it near one edge, 
the disc being placed vertically on an inclined plane which 
is sufficiently rough to prevent sliding. 

This problem is essentially the same as if one side of the 
disc were loaded with a heavier substance, or if, from any 

other cause, the centre of gravity 
is not at the centre of the disc. 

Let g be the centre of gravity 
of the body. The point of sup- 
port for equilibrium must be at 
d vertically under g, which will 
also be a point of tangency of 
the circle and plane. 
With c as a centre, and radius 
eg, describe a circle ; then if the centre cf gravity be at g' 
vertically over d, the body will also be in equilibrium. In 
the former position the equilibrium will be stable ; in the 
latter, unstable. In the latter case, if the body be dis 




Fig. 112. 



[256,257.] CONSTRAINED EQUILIBRIUM. 



173 



turbed, it may appear to roll up the plane, but the centre 
of gravity will really be falling until it assumes a position 
of stable equilibrium. If the vertical through d falls out- 
side the circle gg' s it will not be in equilibrium in any 
position ; if tangent to it, the body will be in equilibrium 
in only one position. 

256. To find the inclination of the plane so that the, 
unbalanced disc of the preceding problem will just be on 
the point of rolling down the plane. 

According to the conditions given in the preceding arti- 
cle, the vertical through d must be tangent to the circle 
gg r . Through c draw a horizontal line, and a right-angled 




Fig. 113. 




triangle cgd, will be formed, in which the angle cdg, Fig 
113, will equal CAB ; hence, 



sin A = sin d = 



eg 
cd 



__ dist. of centre of gravity from centre of circle 
radius of the circle. 

257. A small prismatic bar rests upon two smooth fav 
cMned planes / determine the position for equilibrium,. 
Let AB be the beam, and E its centre of gravity. The 



174 STATICS. [258,259.] 

reactions at A and B must be perpendicular to the respect- 
ive planes : and since these reactions, and the weight W, 
are the only forces which act upon the body, their lines of 
action must all meet in a common point. Hence, the 
perpendiculars AB and BB must meet the vertical 
through E at a common point. If a string ABB be 
attached to the extremities, A and B, of the bar, and 
hung on a smooth pin at B, vertically over the centre of 
gravity E, there will be no tendency to slide on the pin, 
and the bar will remain at rest in the inclined position. 
If the inclinations of the planes are given, a formula may 
be found for the inclination of the bar. 

258. A prismatic bar rests upon the edge of a smooth 
hemispherical bowl, one end being against the inner sur- 
face of the bowl ; required the position for equilibrium. 

Let E be the centre of gravity of the bar, and G the 
centre of the bowl. At A the reaction will be normal to 

the surface of the bowl ; and 
hence, its direction will coin- 
cide with the radius and pass 
through the centre G. At F 
the reaction will be perpen- 
dicular to the bar ; hence, 
when the radius through A 
prolonged, meets the perpen- 
dicular ED, at a point B in a 
vertical through E, the bar will be in equilibrium. 

259. A weight W is on the arc of a vertical circle wnose 
centre is O, and is held in position by a weight P sus- 
pended from a cord, which passes over a phi at G, vertir 
catty over G ; required the position for equilibrium^ 
there being no frictional resistances. 




Fig. 115. 



[ 260.] 



CONSTRAINED EQUILIBRIUM. 



175 



The tension upon the string will be equal to the weigh* 
P ; hence, the body W will be held by a force equal to P 
acting in the direction c G, the force of 
gravity acting vertically downward, 
and the normal reaction of the curve. 
The normal to the circle coincides with 
the radius passing through the point, 
and hence, passes through the centre of 
the circle. 

Draw a vertical ca to represent the 
weight W, and ah, parallel to cO, to re- 
present P ; then the position must be 
such that the extremity b, of the line 
ah, will fall on the normal cO. The 
triangles oho and OOc are similar, hence, we have 

W: Plica: ah 

nOOi Oc; 

,.Oc = ^Oa 




260. A particle W, attached to one end of a string, is 
placed on the convex side of a smooth parabola, and held 
in position hy a weight P attached to the other end of 
the string, the string passing over a smooth pin at the 
focus of the parabola ; required the position for equili- 
brium. 

Let the axis of the parabola be vertical, O the focus, 
and c the position of the weight W. Construct the tri- 
angle of forces cba as before, ce being the normal to the 
parabola. Since ca is parallel to the axis Ce, the norma] 
will bisect the angle Oca ; hence,' 



176 



STATICS, 



[261.1 



Ccb = bca ; 
but, from the construction, 

abc = Ccb ; 
.*. abo = boa ; 
therefore, the triangle abo must be isosceles, and we have 

ab = ac. 



or. 



W=P, 



which establishes a relation between the given quantities. 
It does not determine any particular point, but it shows 




Fro. 117. 




that, in order that the weights shall be m equilibrium at 
any point, they must equal each other, in which case they 
will be in equilibrium at all points on the curve. 

261. Instead of a parabola, let the curve be an ellipse, the 
major axis being vertical and the pin at the upper focus. 
Construct the triangle of forces cab as before, ce being 
the normal. Then, 

W:P::ca:ab 
::de : dc. 



T262.] CONSTRAINED EQUILIBRIUM. 177 

Let e be the eccentricity of the ellipse, then it is shown 
in Conic Sections, that 

de : dc : : e : 1 ; 
hence, 

W:P;:e:l; 
or, 

W 
P= e > 

that is, when the ratio of W to P equals the eccentricity 
of the ellipse, they will be in equilibrium for all positions 
of W on the curve ; and if they have any other ratio they 
will not be in equilibrium at any point, except at the ex- 
tremities of the axis, A and B. 

[In ' ' Conic Sections " the proportion involving e, given in this 
article, may not be given in this form. If a be the semi-major 
axis, and x the abscissa of the point c in reference to the centre, 
then the following equations are usually given, viz. : 
the radius vector, dc, = a ± ex ; 
the distance from the focus to the foot of the normal^ 
de, = e(a ± ex) ; 

the plus sign being used for the distance e'd, and the minus sign 
for ed. These values readily give 



de _ e(a ± ex) _ 
dc~ a ± ex 

hence, the proportion given in the text.] 



262. Let the curve he an hyperbola, the pin being in the 
upper focus. 

Proceeding as before, we find 

W 

but in this case e exceeds unity ; hence FT must exceed P, 
while in the former case TPmust be less than P, 



178 



STATICS. 



[263.1 



263. Suppose, in the preceding case, that the pin over 
which the string passes is at the centre of the hyperbola. 
Let a be the semi-major axis, b the semi-minor axis, the 





axis of x horizontal, and y vertical. The axis of the hyper- 
bola being vertical, the equation will be 



V _ jy = - aW. 



(1) 



Construct the triangle of forces cab t in which ah wiU be 
parallel to cC \ draw cD horizontal, and we have 

W:P::ca:ab 
::Ce: Oc; 

which, according to the principles of Conies, beeorpog 

W:P::e\OD: Oc; 

or, observing that CD — y, and Dc — %, we have 



hence 



W: P'.'.JyiYf+a?; 

W^.P^.ieY'.f+x 2 ; 
which reduced to an equation, gives 

WY+ W 2 a? = P 2 eV. 



(2) 



[263.] CONSTKAINED EQUILIBRIUM. 179 

According to Conies we also have 
« a 2 +b 2 



" a 2 ' 



(3) 



Eliminating a and x from these three equations, we find 



bW 

rz 

b 



y <? 4/ w 2 - e*P 2 ~ GI> 







EXAMPLES. 

1. In Fig. Ill, if TF= 50 lbs., A = 30°, Fof= 60°, what 

must i^be for equilibrium ? 

2. In Fig. 121, if CAB = 45° 

W — 60 lbs., what must F 

be, acting horizontally, for 

equilibrium % (The value of 

F may be found from the 

equation of Article 253, observing that <f> will be 

- 45°.) 

3. Find the value of F, Fig. Ill, for equilibrium when it 

acts parallel to the plane. (Deduce it from the 

equation of Article 253.) 
4u Similarly, find the value of F, for equilibrium, when 

it acts horizontally. 
5. In Fig. Ill, if A = 30°, W= 50 lbs., coefficient of 

friction fi = 0.2, the angle <f> = — 5° ; what will 

be the value of F so as to just prevent motion down 

the plane ; also so that it shall be just on the point of 

moving it up the plane I 



ISO STATICS. [263.J 

6. In the same figure, if A = 45°, W = 100 lbs., p = 

0.15, what will be the value of F, acting parallel 
to the plane, so as just to prevent motion down the 
plane. 

7. In Fig. 113, if the diameter of the small circle equals 

the radius of the larger one, and the remainder of 
the larger one be homogeneous, what will be the 
greatest inclination of the plane A O that the body 
may remain at rest, there being no sliding ? 

8. The weight W being 20 lbs., P 15 lbs., in Fig. 116, 

the radius of the circle 2 feet, the distance O, from 
the centre to the pin, being 4 feet ; required the 
angle O Oc for equilibrium ? 

9. In Fig. 116, if 00 = Oc = the radius of the circle, 

what will be the relation of P and W, and the angle 
OOc, for equilibrium. 

10. In Fig. 120, if W = 100 lbs., P = 25 lbs., b = S feet, 

and e =■ 2, what will be the distance OD for equili- 
brium. 



EXERCISES. 

1. If a force, equal to the reaction of the plane, be substituted for the 

plane, will the body remain in equilibrium ? 

2. If a plane is perfectly smooth, can a body remain at rest on it if the 

plane has any inclination ? If the body were pressed normally 
against the plane, would it remain at rest ? 

3. In Fig. 112, what will be the position of the disc so that the poten- 

tial energy shall be greatest, and what, so that it shall be least ? la 
Fig. 113, is the potential energy a maximum, a minimum, or in- 
different ? 

4. Is the potential energy of the bar in Fig. 114 a maximum or a mini- 

mum ? Similarly, in regard to the bar in Fig. 115 ? 

5. When the relations of P to W are such as to secure equilibrium in 

Figs. 117, 118, and 119, will the potential energy be a maximum, 
a minimum, or indifferent ? 



[263.] CONSTRAINED EQUILIBRIUM. 181 

6. In Article 263, when the bodies are in equilibrium, will the potential 

energy be a maximum or a minimum ? 

7. In Article 262, can the bodies be in equilibrium on the curve, if P 

exceeds W? 
S. In Fig. 118, if W exceeds P, can there be equilibrium at any point 
of the curve ? 

9. In Fig. 115, what is the longest bar that can rest on the bowl and 

have one end, A, in the bowl. 

10. If the planes are equally inclined in Fig. 114, what will be the posi 

tion of the bar when in equilibrium ? 

11. If the plane AG, Fig. 114, were horizontal and perfectly smooth 

what must be the position of the bar for equilibrium ? 



CHAPTER XII. 



ANALYTICAL METHODS. 



264. Analytical Mechanics is generally understood to 
refer to that system of analysis in which the equations of 
equilibrium, or of motion, are established in reference to 
a system of coordinates, and the results obtained by opera- 
tions upon those equations. The coordinates most com- 
monly used are Rectangular or Polar. 

265. Before establishing the general equations, we ob- 
serve that when the forces are in a plane they may be 





Fig. 123. 



reduced to a single resultant, or to a single couple, Article 
194. If they are not in one plane, let P, Fig. 123, be a 
force applied at jB, and F another force applied at A. At 
any point A, of the force 7^ introduce two equal and oppo- 
site forces, each equal and parallel to P, and as they mutu- 
ally destroy each other they will not change the mechanical 
condition of the problem. Now, combining P at B with 
— P at A, we have a couple, whose moment is 

P.AB; 
and the other forces, i^and + P, at A, will have a single 



[266, 26? ] 



ANALYTICAL METHODS. 



183 




Fig. 124. 



resultant. A similar operation may be pei formed when 
there are several forces ; hence, we infer that, when the 
forces are not in a plane, they may be reduced to a single 
force and a single couple. This result, however, may be 
proved from the following equations. 

266. General Case. — Forces may be applied at any or 
all the points of a body, and act in all conceivable direc- 
tions. The angles which the 

action -lines of the forces make 
with the axes may be determined 
as in Article 157. The origin 
of coordinates may be taken at 
any point, and the rectangular 
axes have any position in refer- 
ence to the body or the forces. 
The axis of x will be considered 
as positive to the right, y posi- 
tive upward, and z positive in front of the plane yx. 

267. Forces Resolved. — Let the coordinate axes be 
rectangular, and the forces resolved parallel to them. 

Let F^ F 2 , F 3 , etc., be the forces, 

Oi, 02> #3) e tc., the angles which the lines of the 

forces make respectively with the axis 

OX, 
A* $2> /5 S , etc., the corresponding angles with OT, 
7ij 72) 73) e tc., the corresponding angles with OZ, 
JT, IT, Z, the algebraic sum of the components of 

the forces parallel respectively to the 

axes of x, y, and s, 
2?, the resultant of all the forces, and 
a, b, and c, the angles which it makes with the axes 

as, y } z, respectively. 



184 STATICS. [267.1 

The components of the forces parallel to the axis of x 
will be, according to Article 156, 

F x cos <&!, F 2 cos 02> F$ cos Og, etc., and 
R cos a. 

These components constitute a system of parallel forces, 
which are not confined to one plane, but are simply paral- 
lel to the axis of x. The resultant of any two of the forces, 
since they are in one plane, will be their algebraic sum, 
Articles 190 and 198 ; or. 



F x cos a t + F 2 cos 0%. 

This resultant and a third force will be in one plane, 
and, hence, the resultant of this resultant and a third force 
will be their algebraic sum ; or, 

F x cos a x + F 2 cos «2 -f F s cos 03. 

Continuing this process, we finally have for the result- 
ant of all the components parallel to the axis of x, 

X = F x cos Oj -\-F 2 cos a 2 + F S cos Os + etc. = SFcos a ; 

which value must equal the ^-component of the result- 
ant of all the forces ; hence 

X = B, cos a =2Fcos a. 

Proceeding in a similar way with the y and z • compo- 
nents, we have 

Z = i? cos c = SFcos 7. 
Squaring and adding, we have 
i^^P+P + Z 2 

= (XFoob a? + (SFcob Pf + (XFcos y)\ 



[268.J ANALYTICAL METHODS. 1S5 

The direction of the axis may be determined from the 
equations 

X , Y Z 

cos a = -J* , cos b = -q , cos <? = -^ . 

If there is equilibrium we have 

i?=0; 
.% X=0; r = 0; Z=0. 

These equations are the same as those for concurrent 
forces. Article 161 ; hence, the tendency of a system of 
forces, having given intensities and directions of action, 
to produce translation will be the same whether they con- 
cur or not, and will be independent of the points of appli- 
cation of the forces. 

268. Moments of the Resolved Forces. 

Let 

&1, ?/i, &i be the coordinates of the point of applica- 
tion of F l9 and a corresponding notation for 
the forces F 2 , F s , etc. 

The component of F u parallel to the axis of x, will tend 
to turn the point of application either to the right or left 
about the axis of s (unless it be in the plane a?s), and the 
arm of the component will be y x \ hence the moment will 
be, Article 164, 

F x cos a x . y x ; 

and the component of F x , parallel to y, will also tend to 
turn the point of application about the same axis, s 3 and 
the moment will be 

Fi cos ft . Xi. 

In Fig. 1 25, the component F x cos a, tends to turn the 



JFloos fi 


;>7> 


7h x 


*i 


^fX.x 




; 


^JJe 


? l 





186 STATICS. [268.] 

point of application right-handed, and F x cos fi u left- 
handed ; hence, the resultant moment in reference to the 
axis of z will be 

F x (x x cos J3 X — y x cos a x ). 
The quantity in the parentheses is 
%cosa x the arm of the force F x , as shown 

o\ -11 in Article 176. This expression is 

z equivalent to 

F x cos fi x . x x — F x cos a x . y x ; 
and the other forces give 

F<l cos /3 2 . x% — F 2 cos a 2 . y 2 ; 

and taking the sum of these expressions, we have 

2F cos /? . x — 2F cos a . y ; 
or 

and similarly, in reference to the axis of y, we have 

Xz-Zx; 

and in reference to the axis of x, 

Zy - Yz. 
Let 

d = the moment of the resultant couple, which, as 

shown in Article 171, may be represented 
by its axis ; 
d = the angle which the axis of the resultant 

couple makes with the axis of x, 
e = angle which it makes with the axis of y, 
f— angle which it makes with the axis of z ; 
then will the component of the resultant couple, in refer 
ence to the axis of x, be 

Gcosdi 



[269.] PROBLEMS. 187 

and similarly for the others. Hence, we have for equili- 
brium 

G cos d = Zy — Yz, 

G cos e = Xz — Zx, 
G cos f = Yx-Xy. 

If the given forces are in equilibrium in reference to 
rotation, we have 

G = 0; 
.-. Zy = Yz, 
Xz = Zx- 
Yx = Xy; 

the last of which may be deduced from the two preceding 
by eliminating z. 

In many cases it is advisable to find the moments of the 
forces directly instead of the moments of the components. 



Problems. 

269. A cord %s secured at the points A and C, and sus- 
tains a weight attached to it at B / required the tension 
of each part of the cord, 

the weight of the cord be- t ^^^— - 4?. .^^^^^rt, 

ing neglected, and the ^^~\f^~~ 

parts AB and BC being (T) w 

unequal. FlQ . 126. 

Let t = the tension of AB, 

ti = the tension of B C, and 
W = the weight of the body. 

Taking the origin of coordinates at any point as B, 
v horizontal, and y vertical, and resolving the forces 
horizontally and vertically, we have 



188 STATICS. [270,] 

X = t cos ^J.Z> - £ cos BOD =: 0, 

r = t sin A4Z> + t t sin ^6/Z? — TF= 0. 

There being only two unknown quantities, the equation of 
moments will be unnecessary. If the origin of momenta 
be at B, all the moments will vanish. 

From the first of these equations we have 

t cos BAD = ti cos BOD ; 

hence, the horizontal components of the tensions are equal 
to each other. From the last equation we have 

, _ cos BOD , 
cos^iT 15 

which, combined with the second equation, gives 

_ sec BAD w 

tun BAD+tsm BOD ' 

sec BOD 
1 " tan BAD + tan BOD 

270. A weight W is attached to a cord DB of given 
length and pushed from a vertical wall by a strut BA of 
known length / required the tension, t, 
of the string, the compression, c, of the 
strut, and the upward push, F, on 
the vertical wall, when there is equili- 
brium. 

Let AD be known, and the angles of 
the triangle ABD computed. 

Take the origin of coordinates at A, 
x horizontal, and y vertical, and posi- 
tive upward. Resolving the forces acting at B, we have 
(see Article 157), 



D 


V 




zl 








*/\ 


*^B 


J!^ 


( 
F 


Aw 



[271.] PROBLEMS 189 

X = t cos (90° + $) + c cos (270% 6) + IF cos 270°= ; 
T= t sin (90° + <£) -f tf sin (270° + 0)-!- TFsm270° = 0; 
and taking the moments of the forces directly, we have 

2Fa = + t.AC+c.0-W.BF=0. 
These reduced, give 

— t sin <j) •+- c sin 6 = ; 

+ t cos — c cos — W= ; 

*. ^L#sin - TT. 4J?sin (9 = 0. 

AC 
These equations solved, observing that sin = -jj,, and 

. a EB . 
Bm AB' glve 

AD sm</> ^. (7 ' 

In a similar way, resolving the forces c, F, and X at 4, 
and we find 

AD ' 
and EB 

AD 

271. Two braces, or rafters, secured at their lower ends 
by a horizontal tie rod, carry a weight W at the point 
where they meet ; required the vertical action at the sup- 
ports, the tension of the horizontal tie, and the compres- 
sion of the braces. 

First, consider the parts as rigid, and determine the re- 
lations between the external forces. 

Let the braces be equally inclined, then will D, verti- 



190 



STATICS. 



[271.J 



cally under W, be midway between A and B. Also, let 
V x be the vertical action of the support at A, and V that 
at B. Take the origin of coordinates at A, x horizontal 
and y vertical, and we have, from Fig. 128, 

X = V, cos 90° + TFcos 270° + Fcos 90° = ; 
7 = V 1 sin 90° + IT sin 270° + Fsin 90° = ; 
2Fa = V 1 .Q-W.AI) + V.AB = 0; 



and these give 



X=0; 



W=0 



2Fa = V.AB-iW.AB = 0; 
which readily give 

v=\w=v„ 

hence, each support sustains one-half the load W; a result 
readily deduced from the principle of the lever. 




Fig. 128. 




Secondly, to find the stresses which are transmitted 
along the pieces, conceive that a plane EF is passed 
through two of the pieces, cutting them at the points E 
and F. Let the part AEF, between the end A and the 
plane section, be removed, and let the forces c and t, 
equal to the compression and tension which previously ex- 
isted, be substituted. The frame in the new condition 
will be in equilibrium, the forces c and t being the unknown 



[272.] PEOBLEMS. 191 

quantities to be determined from the equations of equili- 
brium. The direction of action of these forces is repre- 
sented by the arrow heads, Fig. 129. 

The origin of coordinates remaining at A, we have 

X = Fcos 90° + TTcos.270° + t cos 180° 

+ ccos CAD = 0, 
Y= Fsin 90° + W sin 270° + tsm 180° 
+ c sin CAD = 0, 
SFa = V.AB- W. AD+t.O + c. = 0; 
which give, observing that V = -J W, 

— t + ccos CAB = 0, 

— 'iW+own CAD = 0; 
and these give 

o — i TTcosec CAD, 
t =iWcotCAD; 
or, 

c cos CAD = t, 
c sin CAD =iW; 

from which it appears that the horizontal component of 
the compression on A C equals the tension of the horizon- 
tal tie AB, and 

The vertical component of the compression on AC 
equals the vertical action at A. 

In a similar way the stresses on any frame may be de- 
termined, provided the plane section does not intersect 
more than three acting members. There being three 
equations for equilibrium for forces in one plane, two for 
forces, and one for moments, only three unknown quanti- 
ties can be determined from them. 

272. Remark. — These solutions are given that the stu- 
dent may learn the process of establishing the general 
equations, and not because the solutions are the shortest 



192 



STATICS. 



[ 273.] 



or the most simple The solutions here given are appar- 
ently longer than are necessary, for those quantities and 
terms which reduce to zero need not be entered in the 
equations ; but it was thought best, for practice, to enter 
every force in each equation. The Analytical method is 
not used because it makes the solution of a particular 
problem shorter than by special methods, but because it 
establishes a uniform method of procedure, and also fur- 
nishes a powerful method of investigation in more diffi- 
cult problems. 

273. The preceding problem may be solved by the 
following shorter method. Let Cb represent the weight W, 

and ba be drawn parallel to BC, 
and ad horizontal ; then will ab 
represent the compression on 
CB, aC = ab = the compres- 
sion on CA, and ad the tension 
on AB. Cd=db = -J TT; hence, 
we have 
ad = Cd cot Cad ; 

i Wcot CAB ; 

a C = Cd cosec Cad, 

c = i W cosec CAB. 




Fig. 130. 



or, 



and 



or, 



t 



EXAMPLES. 

1. In Fig. 126, if W= 100 lbs., AC =5 ft., AB = BO 

= 3 ft., what will be the tension of the cord ? 

2. In the same figure, if the length of the cord is 10 ft., 

AC, 5 ft., what must be the length of AB that the 
tension of it shall be twice that of BG\ 



[273.] PROBLEMS. 193 

3. If AB be horizontal, in Fig. 126, and BC inclined, find 

the tension on AB and BC. 
L In Fig. 128, if AB is 20 ft., BO 10 ft., and W 500 lbs., 

find the tension on AB and the compression on A C. 
5 Required the inclination of the rafters, so that the stress 

on each will equal the weight W. 
6. Required the inclination of the rafters so that the stresa 

on AB shall equal the weight W. 



EXEKCISES. 

1 In Fig. 122, if the resultant of F and F x equals R, and another force 
equal to R be introduced at the point of concurrence, and another 
force equal and opposite to R acting on another part of the body- 
be also introduced, will the four forces have a single resultant ? 

2. If the force R, at the concurrence of F and Fi be removed, will the 

three remaining forces have a single resultant ? 

3. In Fig. 125, if the point of application of the force be in the second 

angle, both components will turn the point of application right- 
handed ; does the expression for the moments need to be modified 
for this case ? 

4. In Fig. 126, can the cord be drawn into a straight line when the 

weight B is upon it ? 

5. If the parts of the cord AB and BG are vertical, what will be the 

tension on each part ? 

6. If the points of support A and C are not in the same horizontal, while 

the inclination of the parts AB and BC remain the same, will 
the tension remain the same ? 

7. If the weight be moved to and fro on the cord, in what curve will be 

the locus of the point B ? 

8. In Fig. 128, if the lower ends of the braces, A and B, be moved to- 

wards each other, will the compression on the braces be increased 

or decreased ? Will the thrust at the lower ends be changed ? 
. What will be the effect upon the stresses if the apex be lowered by 

lengthening the cord AB? 
0. If a vertical strut CD be placed under C, will there be any stress on 

the rafters ? 



CHAPTER XIII. 



STRENGTH OF BARS AND BEAMS. 



274. Strength of Prismatic Bars. — It has been ofc 
served, Articles 129 and 130, that, if a solid be pulled in 
the direction of its length, it will be elongated. We also 
know that if the pulling force be sufficiently great, the 
bar will be broken. 

The strength of solids is determined by experiment. 
The most common way of doing it is to take a bar whose 



rO 







Fig. 131, 

cross- section is somewhat larger than the section which it 
is desired to test, and then turn down a portion near the 
middle of .the piece to an exact cylinder of definite dia- 
meter. The piece is then placed in a powerful machine 
and pulled asunder. The total pulling force being mea- 
sured, the stress per unit of section is easily computed. 
By experimenting upon pieces of different diameters and 
lengths, and with different materials, certain general facts 
or laws have been determined. 



1275,276.] STRENGTH OF BARS AND BEAMS 195 

275. Results. — The strength of prismatic bars varies 
directly as their cross-section. This is the law assumed 
in practice, and it appears to be nearly evident without 
experiment ; but, like all other physical laws pertaining 
to solids, it is not rigidly exact. It is found that, when 
the reduced portion ab is very short, the piece is stronger 
than when it is longer, but no law for this increase of 
strength is known. Bars of small cross-section generally 
appear to be stronger in proportion than those of larger 
cross-section, but the difference may be more apparent than 
real, for it may be due to the difficulty of producing a pull 
exactly in line with the axis of the larger piece, and cross- 
strains will generally break a piece more easily than a 
direct pull. If, however, the pieces are made small by 
being so forged before they are turned, instead of being 
turned down from larger pieces, the difference in strength 
may be due to the process of manufacture. 

When a piece is elongated, its cross-section is contracted. 
When the material is ductile, the contraction may be con- 
siderable before fracture takes place. Thus, in some ex- 
periments upon wrought iron, the cross-section has been 
reduced in this way to one-half of its original area. 
When the part ab. Fig. 132, is very short, the contraction 
is much less than when it is longer. The shoulders of the 
larger portion appear to aid in resisting the contraction 
of the smaller part, and this may account for the increase 
of strength above referred to. 

276. Modulus of Tenacity. — The pull, in pounds, 
which will break a bar whose cross-section is one square 
inch, is a measure of the tenacity of the material and is 
called The Modulus of Tenacity. It is represented 
byT. 



196 STATICS. [277-279. J 

Mean Values of T. 

For wrought iron 45,000 lbs. to 60,00C lb& 

For steel 70,000 " to 190,000 " 

American cast-iron 20,000 " to 45,000 « 

English cast-iron 13,000 " to 25,C00 rt 

Ash. (English) : 15,000 " to 17,000 " 

Oak {English) 9,000 " to 15,000 « 

277. Formula for the Tenacity of Solids. 

Let 

T = the modulus of Tenacity, 
P = the pulling force, 
S = the area of the cross-section ; 

then we have 

P = TS. 

278. Strength of Beams. — In order to make a formula 
for the strength of beams, it is necessary to know the law 
of action of the forces within the beam. Within the elas- 
tic limits this law is quite perfectly known, but, when a 
state bordering on rupture is reached, the law of action is 
quite complex, and is not accurately known. But the same 
law of resistance is assumed for both cases, and the error, 
if any, which results from the assumption is corrected 
by means of a coDstant factor. 

279. Law of Resistance. — When a beam is fixed at 
one end and loaded at the free end, as in Fig. 133, the 
fibres on the upper side are elongated and on the lower 
Bide, compressed, and there is a surface between the ex- 
tended and compressed elements which is neutral, and 
is called a neutral surf ace. It is assumed that the stress 
on the fibres varies directly as their distance from the 



1280-281.] STRENGTH OP BARS AND BEAMS, 



197 



neutral surface. The intersection of the neutral surface 
with a vertical, Longitudinal plane is called the neutral axis* 




Fig. 133. 



280. Modulus of Rupture. — The Modulus of Rup- 
ture is the stress at the instant of rupture on a unit of 
area of the cross-section which is most remote from the 
neutral axis. It is represented by R. 

281. Expression for the Moment of Resistance of 
Rectangular Beams. — In Fig. 133, let pr be the neutral 
axis. Take In = mo = ab = dc = R. Pass the planes 
po and pc, then will the wedge pq-noml represent the 
total pulling stress, and pq-abcd the total pushing or 
compressive stress. 

Let 

o = Im = the breadth of the beam, and 
d = la = its depth ; then 
%od=. axeaplniq, and 
\Rhd = the volume of the upper or lower wedge. 
The moment of this resistance equals the total resistance 
into the distance of the centre of gravity of the wedge 
from the neutral surface. The centre of gravity of the 



198 STATICS. 

wedge is at the same distance from the neutral surface aa 
that of the triangle j?nl; that is \j>l\ hence, the arm of 
the resistance is \jpl — \^\d = \d and the moment is 

iMd.id = ^md 2 ; 

and the total moment of resistance of both the upper and 
lower stresses will be twice this value, 01 

\Rbd\ 

This expression is true for all rectangular beams whosi 
sides are vertical, whether the beam be fixed at one end, 
supported at its ends, or otherwise, and whether the beam 
be loaded at one point, several points, or uniformly loaded. 

282. Problems. — 1. Let a rectangular beam he fixed at 
one end, to find a load P, placed on the free end, which 
will just break the beam. 

In Fig. 133, let l=pr = the length of the beam ; then 
the moment of P, in reference to the fixed end as the 
origin of moments will be 



PI; 






and this must equal the moment of stress ; 


hence, 




PI = \Rbd 2 






•P-P^ 


• 


a) 


From this equation we find 






M ~bd 2 ' 


. 


(2) 



from which the value of R may be easily computed when 
the breaking weight P and the dimensions of the beam 
are known. The value of R has been determined for a 
variety of substances, and the results given in tables in 



[282.J STRENGTH OF BARS AND BEAMS. 199 

works on the Eesistance of Materials. It is a remarka- 
ble fact that the value of R is not the same as T for any 
substance. 

Equation (1) is true for any stress less than that which 
will break the beam. It is customary, in practice, to use 
a fractional part of R, called the factor of safety, when 
beams are to be proportioned. About \ to -J of R is used 
for wrought iron, ^ for wood, J to -J- for cast iron, 
which giye for the safe value of R per square inch about 

10,000 lbs. to 12,000 lbs. for wrought iron, 
1,000 lbs. to 1,200 lbs. for wood, 
5,000 lbs. to 6,000 lbs. for cast iron, and 

12,000 lbs. to 20,000 lbs. for steel. 

Any one of the quantities of equation (1) may be found 
when all the others are given. 

2. Let the learn he fixed at one end and uniformly 
loaded. 

Let W equal the total load ; then will the lever-arm of 
the load be the distance from the fixed end B to the centre 
of gravity of the load, which 
is \l ; hence, the moment is 

im, 

and the equation for equili- 
brium becomes 

iWl = %Rhd*. Fia ' m ' 

3. Let the learn he supported at its ends and loaded in 
the middle. 

The point where a beam is most liable to break is called 
the dangerous section. If the beam at this section is 
strong enough to support the load the beam will not break. 
In this problem the dangerous section is at the middle of 




200 STATICS. [282.] 

the length; hence, the equation of moments should be 
established for this point. 

The reaction of each support will be \P, and the arm 
of this force, in reference to the middle point of the beam 
as an origin of moments, is il } and the moment is 

and the equation becomes 

\Pl = \Rbd\ 
4. Let the beam he supported at its ends and uniformly 
loaded. 

Each support will sustain one -half the load, or 

xw~ 

2 "* 



M 




D 

Fig. 135. Fig. 136. 

which acts upward and tends to turn the beam one way. 
The arm of this force is \l in reference to the centre of the 
beam taken as the origin of moments, and its moment will be 

The load between the middle of the beam and one of the 
supports, which is also -§■ W, acts downward, tending to 
turn the beam in the opposite direction to that of the 
former force ; and the arm of this force is the distance 
from the origin of moments to the centre of gravity of 
this load, which is \l ; hence its moment is —i W.^l; and 
the resultant moment will be 

iWl-iWl = iWl; 
and the equation for equilibrium becomes 
iWl = %Rbd\ 



1282.] STRENGTH OF BARS AND BEAMS. 201 

EXAMPLES. 
[It is best to reduce all the linear dimensions to inches.] 

1. A rectangular beam, whose depth is 8 inches, length 8 

feet, R = 1,400 lbs., is supported at its ends; re- 
quired the breadth so that it will carry 500 lbs. per 
foot of its length. Am. b = 3^ inches. 

2. A rectangular beam is fixed at one end and is required 

to carry 1,000 lbs. at the free end ; if I — 8 feet, 
R = 1,200 lbs., and the depth is four times the 
breadth, required the breadth and depth. 

3. A rectangular beam, whose length is 12 feet, breadth 

2 inches, modulus of rupture 10,000 pounds, is sup- 
ported at its ends ; required the depth so that it will 
support 8,000 pounds placed at the middle of the 
beam. 

4. If a beam whose length is 10 feet, breadth 4 inches, 

and depth 9 inches, is supported at its ends, and 
broken by a weight of 20,000 pounds placed at the 
middle of its length, what is the value of R ? 

5. A beam whose breadth is 1-J- inches, depth 3^- inches, 

is supported at its ends and loaded at the middle 
with 10,000 pounds ; what must be its length so 
that the stress on the outermost fibres shall be 20,000 
pounds ? 

6. A beam whose length is 15 feet, breadth 6 inches, and 

depth 12 inches, is supported at its ends ; required 
the load, uniformly distributed, which it will safely 
support, calling the safe value of R 12,000 pounds. 

7. An iron beam, two inches square, projects horizontally 

from a wall ; what must be its length to break it- 
self at the wall, the value of R being 30,000 pounds, 
and the weight of the material being J pound pei 
cubic inch ? 



CHAPTER XIY. 



MOTION OF A FARTICXE ON AN INOLINEI PLANE. 

283. To find the formulas for the movement of a parti- 
cle down an inclined plane by the action of gravity, all 

resistances being neglected. 

Let 

A O be the plane, 

W = the weight of the particle, 

F= the component of the force of gravity par- 
allel to the plane, 

s = the distance over which the body moves in 
the time t, 

v = the velocity acquired in the time t, and 

</> = CAB = the angle of elevation of the plane. 

The force of gravity resolved parallel to the plane gives 

c F— TTsin <f>, 

which, being constant, will, ac- 
cording to Article 60, produce a 
constant acceleration, the value of 
which may be found by Article 




Fia. 181. 



86, and is 



- F W sin cj> 



1284.] MOTION OF A PARTICLE. 203 

This value oif in the equations of Article 24, gives 
v = gt sin <£, (1) 

« = V2^ sin <£, . . . (2) 
* = \g# sin 0, . . . (3) 

W-.. . . (4) 

<7 sin 9 

Let 8 = A C, then, from the figure, we have 

ssin (f) = BO, 
which, substituted in equation (2), gives 



■v 



*> = V2g.BC\ 
which equals the velocity acquired in falling through a ver- 
tical height BO. Since a similar result will follow from a 
plane having any inclination, and hence, from several suc- 
cessive planes having different inclinations, it follows that 
the velocity acquired by a particle in falling from one 
point to another ivhen there are no resistances, will be in- 
dependent of the path over which it moves, and will equal 
the velocity acquired in falling vertically through a height 
equal to the height of one point above the other. 

This result also follows directly from the principles of 
Kinetic Energy, for the energy imparted by gravity equals 
the weight of the body into the height through which it acts. 

The formulas of this article may be deduced from those 
of Article 247, by supposing that one of the bodies van- 
ishes. 

284. Initial Velocity. — If the body is projected down 
the plane, or, in other words, has an initial velocity v at 
the instant t begins to be reckoned, we have, 
v = v + gt sin $, 
8 = v Q t + igt 2 sin </> ; 
& = vf-r %gs sin <f>, 



204 



KINETICS. 



[285,286.: 



and, if the body be projected up the pkne with a velocity 
# , we have 

v = v — gt sin $, 

s = v t — \gft sin <j>. 



% 



2 _ 9 



vas sin 



Problems. 

285. The times of descent down all chords of a vertical 
circle, which pass through either extremity of a vertical 
diameter, are the same. 

Let AGB be the circle, and AG any chord drawn 
through A. According to equation (4) of Article 283, the 
time of descent down A G will be 




=v; 



2 AG 



gsin <f> 

Draw GB, then will the angle 
A GB be right, for it is inscribed in 
a semicircle. The angle cf> is the com- 
plement of GAB ; hence, 

AG 

sin <f> = cos A = -rw, 

which, substituted in the preceding equation, gives 

/2AB 

which is not only constant, but is the time of falling 
vertically through the diameter AB. 

286. Find the straight line from a given point to a 
given inclined plane, down which a body will descend in 
the least time. 



[287 J 



MOTION OF A PARTICLE, 



205 



Let A be the point and B G the plane. Through A draw 
a vertical line A 0, and on it take at such a point that 
a circle described with the radius OA shall be tangent to 
BG\ then will the chord AD, drawn from the upper ex- 
tremity A of the diameter, to the point of tangency D, be 
the required line. For, all other lines drawn from A tc 
the right line will be secants of the circle, and, accord- 
ing to the preceding article, the times of descent down 





the internal portions will be the same, and hence, the time 
of descent down the whole line will be greater than that 
down AD. 



[To find the centre of the circle, draw a horizontal line AB, 
which line will be tangent to the circle, and bisect the angle B by 
the line OB ; the point of intersection, 0, of the lines A and 
BO, will be the centre required. From the property of tangents, 
BA will eqnal BD.] 

287. To find the straight line of quickest descent from 
one circle to another, the centres of the circles being in the 
mme vertical line, and one tangent to the other internally 

Let A be the point of tangency. Draw any secant ADC. 



206 



KINETICS. 



[230.] 



and draw BC. On aB as a diameter, draw the circle 
zbB, and draw the chord ab. The angles 6 r and b will be 
right, for each is inscribed in a semicircle, and ab will be 
parallel to DC. The time of descent from DtoC will be 
the same as from a to b, which is the same as from a to B ; 
hence, the time of descent from any point of the circle 
whose centre is 0, down the external portion of the secant 
drawn from A through a point D to the external circle, is 
constant and equal to that down the vertical aB. If D 
is a given point on the circle ADa, then D O will be the 
line of quickest descent. If the common tangent is at the 
lowest point of the circles, at B, and C the given point ; 
draw the secant BO, and Ob will be the required line. 

288. To find the straight line of quickest descent from 
a given circle to a given, line. 

Let AEB be the circle, and OD the line. Through A 
draw the horizontal line A C, make OB equal to A C, and 
draw AD ; the line required will be ED. 




Fig. 13ft. 



289. To find the line of quickest descent from a straight 
line to a given circle. 

Let CD be the line and AEB the circle. Draw the 
horizontal tangent AC; make OD=AC; draw D A, and 
DE will be the required line. 



L 290., 291.] MOTION OF A PARTICLE. 207 

290. Find the time of descent down a rough inclined' 
plane. 

Let the forces be resolved as in Fig. 131, and ft be the 
coefficient of friction. The normal pressure will be 

JST= PTcos^, 
and the resistance due to friction will be 

/i TFcos <f> ; 
hence, the effective moving force will be 
TTsin $ — fiW cos <f> 
= (sin (j> — /jl cos (j>) W\ 
and the conrtunt acceleration will be 

/= (sin£-#cos<%; 

which, substituted in the equations of Article 24, gives 

v = (sin 4>— fjb cos <f>)gt, .... (1) 

v = |/2(sin </> — /*, cos <fygs, ... (2) 

s = J(sin <j> — fi cos 4>)g&. ... (3) 

291. Approximate Formulas. — Substitute the values 
of the sine and cosine in the preceding value of /j and we 
have 

AB\ 
f=\-A7T-V>-T7r)9\ 



/BO _ AB\ 

\AO ^AO/ 



but, when the angle A is small, AB may be considered as 
equal to AC, and BO -4- AC will be the grade (a term 
which is in common use in Railroad Engineering). Let 7 
be tb 3 grade, and the formula becomes 

an<* ve have 

v = ( y- fi)gt, (1) 

v=V2(y-fi)gs, .... (2) 

8 = Hj-fi)gf. (3) 



208 KINETICS. [2921] 

292. Formulas adapted to the movement of cars on 
planes of small inclination. 

The grade is commonly given in feet per mile. Let h 
be the elevation per mile, then 

_ h 

7 ~ 5280 ' 

and the friction may be taken at 7.5 lbs. per ton gross ; 
hence, 

** 2240 ™ 300 ' 
and the formulas become, with sufficient accuracy, 

•=5^(A-lT.«ft • • • (1) 
» = IV(/t-17.6>,. . . (2) 

«= g i 1 (A-17 i 6)#. . . (3) 

If the body has a velocity, the distance it will move on 
a horizontal plane, in being brought to rest, may be found 
by making h — o, in the second of these equations, and the 
time of the movement, from the first, after making h = o 
in that equation ; hence we will have for this case 

t = 9±v (nearly), .... (4) 
s = 4:.6v 2 (very nearly). . , (5) 



EXAMPLES. 

1. A smooth inclined plane is 100 feet long; what must 
be its inclination that the time of descent of a parti 
cle down it shall be 5 seconds ? 



[292.] MOTION OF A PARTICLE. 209 

2. A body is projected up a smooth plane whose slope (that 

is its inclination with the horizontal) is 45 degrees, 
with a velocity of 50 feet per second ; find its posi- 
tion at the end of 3 seconds; five seconds; ten 
seconds. 

3. A body starts from rest at the top of an inclined plane 

whose length is 100 feet, and height 20 feet ; with 
what velocity must a body be projected up the plane 
from its foot in order to meet the former one at the 
middle of the plane % 

4. Find the line of quickest descent from a point with- 

out a circle to the circumference of the circle. 

5. A train of cars starts from rest on an inclined plane 

and runs down it by the force of gravity only ; the 
grade being 40 feet to the mile and the coefficient of 
friction -g-J-g-, what will be its velocity at the end of 
one mile % 

(x A car starts from the upper end of an inclined plane 
whose length is one-half of a mile, and grade 50 feet 
per mile, coefficient of friction ^-J^, and runs down 
the plane ; how far will it go on a succeeding hori- 
zontal plane before being brought to rest, the coeffi- 
cient of friction remaining constant ? 

7. In the preceding example, required the time of the 
movement down the plane; also the time on the 
horizontal plane. 

8 Required the height of a plane which is 1,200 feet long, 
so that a car starting at rest from the upper end of it, 
and running down shall go just 1,000 feet on a suc- 
ceeding horizontal plane, the coefficient of friction 
being T fo. 

9. Suppose that there is an available height of 25 feet for 



210 KINETICS. [292.J 

an inclined plane, required its length so that a body 
descending it shall go just 800 feet on a succeeding 
horizontal plane, the coefficient of friction being ^-g- 

[The examples from 5 to 9 inclusive may be solved by the for 
mulas of the last article. The resistance of the air, and the 
effect of the rolling mass in the wheels, is neglected. In the 
formulas of the preceding article, t is in seconds, v in feet per 
Beoond, a id s in feet.] 



CHAPTEK XV. 

PROJECTILES. 

293. A Projectile is a body thrown into space and 
acted upon by gravity. It is important in the art of Gun- 
nery ; but the solution of problems in that Art pertain- 
ing to the flight of the projectile, to be of any value, in- 
volves the resistance of the air, and this element so com- 
plicates the problem that its solution will not be attempted 
in this work. Some interesting properties may, however, 
be easily determined by assuming that the body moves in 
a vacuum ; and it will be found that the formulas thus 
deduced are of practical value in the science of Hydro- 
dynamics. 

294. The path of a projectile is the line which it de- 
scribes. This assumes that the body is a particle, but if 
the size be considered, we would say that it is the line de- 
scribed by the centre of its mass. 

The horizontal range of a jyrojectile is the distance 
A B, Fig. 139, from the point A, where the body is pro- 
jected, to the point B, where the path crosses the horizon 
tal plane passing through A. 

295. The path described by a projectile in a vacu- 
um is a parabola. 

Let A be the point from which the projection is made, 
AB the line of projection, and v the velocity of projection. 
In a certain time, t, the body will have moved a certain 
distance, AB, under the action of the impulse only; 
hence, 



212 KINETICS. [996.] 

AB = vt 

In the same time it would have moved a distance AD =• 
BO under the action of gravity only ; or 

BO=igA 

When both motions are simultaneous, it will be at the 
point C, at the end of the time t. Eliminating t from 
these equations, gives 

A&J^AD. 

9 

Let h = the height through which a body must fall in 





Fig. 137. 



Fio. 138. 



vacuo to acquire a velocity v ; then, according to Arti- 
cle 72, 

v 2 = 2gh, 

which, substituted in the preceding equation, gives 

AB 2 = M.AD; 
which is the equation of a parabola. The line AB is a 
tangent to the curve, and AD is a diameter. 

296. To find the velocity at any point of the path. 
Let A, Fig. 138, be the point. The velocity will be 
the same as if the body had fallen a height h ; but 4A, 



297,298.] 



PROJECTILES. 



213 



according to the last equation, is the parameter of the 
parabola to the diameter AD. Let EG be the directrix 
of the parabola, then will the velocity be equal to that 
due to a fall through a vertical height EA ; or, 

v=V2g.EA. 

297. To find the position of the focus. Let F be the 

focus, and AL be drawn horizontally. According to a 
property of the parabola, the tangent AB makes equal 
angles with the diameter AD and the line AF drawn from 
the focus to the point of tangency ; hence, 

BAF= JAD 

= EAB = 90°-BAZ; 

therefore the angle between the tangent to the path, and 
the line from the point of tangency to the focus, is the 
complement of the angle of elevation of projection. 

The distance AF— EA, is, according to the preceding 
article, 

v 2 

hence, when the velocity of projection and the angle of 
elevation are given, the focus can be found. 

298. To find the equation 
of the path when referred to 
rectangular axes. 

Let A be the point of pro- 
jection, and P the position of 
the body at the end of a time t. 
Take the origin of coordinates 
at A, x horizontal, and y vertical 
coordinates of P being 




Let a = EAF The 



f 



214 KINETICS. L299-301.J 

AF=x, FP = y y 
we have, 

AF= vt, 

AF= vt cos a = x, * . (1) 

^£7^ = vt sin a, 

.\F.P=vtsma-igfi = y ... (2) 
Eliminating 2 from equations (1) and (2) gives 

which is the required equation. 

299. To find the range AB. Making y = in equa 
tion (3), we have 

x = 0, 
and 

x = 4A cos a sin a = 2A sin 2a = J.i?. 

300. To find the Time of Flight. Substitute the 
value of AB, given in the preceding article, in equation 
(1) of Article 298, and we find 

4Ji sin a v sin a 

t = == 2i • 

v g 

301. To find the greatest height in the path. — 

It will be vertically over Dythe middle point of AB. 
From Article 299 we find 

AB = 2h cos a sin a ; 

which, substituted for x in equation (3) of Article 298, 
gives 

CD = h sin 3 a. 



302-304.] 



PROJECTILES. 



215 



302. Greatest Range. — To find the angle of elevation 
of the projectile which will give the greatest range. 

This requires that the value of AI>, in Article 299, 
shall be a maximum ; hence, 

sin 2a = 1 
.'. a = 45°. 

303. Greatest Height. — To find the angle of elevation 
of projection which shall give the greatest height. This 
requires that the value of CD, Article 301, shall be a 
maximum; hence, 

sin 2 a = 1 
/. a = 90° ; 

that is, the projection must be vertical. 

304. Equal Ranges. — Since the sine of an angle equals 





Fig. 141. 

the sine of its supplement, the value of AB, Article 299, 
will be the same 

for sin 2a, as for sin (180° — 2a) ; 
hence, the range will be the same for the angles 

a, and 90° -a; 
that is, if the angles of elevation of two projectiles be tne 
complements of each other, and have the same initial 
velocity, their ranges will be equal to each other. 



216 KINETICS. [305,306.] 

305. Range on an Inclined Plane. — If an inclined 
plane, OB, passes through the point of projection, it is 
required to find the range on the plane. 

Let the angle BO G— 0, then will 

OB = y = x tan <f>, 

and this value of y, substituted in equation (3) of Article 
298, gives 

~ ~ 2v 2 cos a sin (a — <j>) 

x— 00= j — ; 

gcos<l> 

therefore, 

OB = OCbbo $ = 2 ^ cosasi ° (—»> . 

g cos* 5 9 

When the projectile is moving in a vacuum, gravity is the 
only force which acts upon it. 

306. Problems. — 1. Find the equation of the path 
when the body is projected horizontally. 

Y In this case a = in equation (3) of 

Article 298 ; hence we have 

o? = —4Jiy. 

If y be taken positive downwards, we 
have 

t? = tfiy. 

2. When the body is projected horizontally, find the 
range BO on a horizontal plane below the point of p7*ch 
jection, and the time of flight. 

Let AB = y x . Substitute y x for y in the preceding 
equation, and solve for x, and we find 

BO=x = ^Vky l . 
The time of flight will be that necessary to move the 




r306.] PROJECTILES. 217 

horizontal distance J30 with, the velocity v. Let t be the 
time, then 

vt = J30=2Vh^ 1 
2V% 

• • — • 

V 

3. To find the velocity and angle of elevation so that a 
projectile shall pass through two points whose coordinates 
are known. 

Let Xx and y x be the coordinates of one point, x% and y % 
the coordinates of the other, and these substituted for 
x and y respectively in equation (3) of Article 298, give 

x? 

y l = x l t2ma— — r - 

9 4th cos 2 a 



2„ > 



X2 2 



y 2 = X2 t an a — -^ r - 

9 4Acos 2 a 

Eliminating h gives 



^wp% — #1) 
Eliminating tan a, gives 

h - ^fe — #i) . 
" 4cos 2 a(2/ 1 aj 2 -2/ 2 aJ 1 )' 

hence, 



s 2 cos a y 2/^2 — y&i 



EXAMPLES. 



1, A body is projected with a velocity of 5g at an angle of 
elevation of 45 c tc the horizon ; determine the 
range and greatest height. 



218 KINETICS. [306.J 

2. A Dody is projected horizontally with a velocity equal 

to that acquired by a body falling through a height 
of 15 feet ; required the range BG, Fig. 142, on a 
plane 12 feet below the point of projection. 

3. The horizontal range of a projectile is 1,000 feet, and 

time of flight 15 seconds ; required the angle of 
elevation, velocity of projection, and greatest alti- 
tude. 

Arts, a = 74° 33' 9". 
v = 250.29 feet. 
A = 904.69 feet. 

4. A body projected at an angle of elevation of 45° has a 

horizontal range of 25,000 feet ; required the velocity 
of projection, the greatest altitude, and time of flight. 

Am. v = 896 feet. 
A = 6,250 feet. 
t = 39 + seconds. 

5. The horizontal range of a projectile is four times its 

greatest height ; required the angle of elevation. 

6. A body is projected from the top of a tower 150 feet 

high, at an angle of elevation of 45°, with a velocity 
of 75 feet per second ; required the range on the 
horizontal plane passing through its foot. 

7. The eaves of a house are 25 feet from the ground, and 

the roof is inclined at an angle of 30° to the horizon, 
the ridge being 32 feet from the ground ; find wheiB 
a smooth sphere sliding down the roof will strike the 
ground. 

8. A body passes through two points whose coordinates 

are x x = 400 feet, y x — 60 feet, x 2 == 600 feet, and 
y 2 = 50 feet ; find the velocity, angle of elevation, 
horizontal range, and time of flight. 



[306.] PROJECTILES. 219 

The effect of the resistance of the air may be illus- 
trated by the fact that a certain ball being pro- 
jected with a velocity of 1,000 feet per second in the 
air, its range was found to be about 5,000 feet, instead 
of 31,250 feet, as it would have been in a vacuum. 



EXERCISES. 

1. A body is projected vertically upwards ; what will be its range ? 

2. A ship is sailing due east at the rate of 5 miles per hour ; if two 

bodies are projected from the ship at the same angle of elevation, 
one in an easterly direction and the other in a westerly direction, 
will the range of the projectiles be the same ? 

3. If a ship sails at the rate of 15 miles per hour, and a body be thrown 

from it in an opposite direction with a velocity of 11 feet per 
second, required the actual velocity of projection. 

4. Considering the rotation of the earth, will the range of a projectile 

be the same if it be fired in a westerly direction that it will if 
fired in an easterly direction ? 

5. If a projectile be struck horizontally, when at its highest point, by 

another body, will it reach the horizontal plane in the same time 
as if it had not been struck ? 

6. If two particles be projected from the same point with the same 

velocity, but the angle of elevation of one is as much above 45° as 
the other is below it, will their ranges on a horizontal plane be the 
same? 

7. If two bodies are projected from the same point on a smooth hori- 

zontal plane, with different velocities, and in different directions, 
show that the straight line which joins them always moves paral- 
lel to itself. 



CHAPTER XYI. 

CENTRAL FORCES. 

307. A Central Force is one which acts directly to* 
wards or from a point. The point is called the centre of 
the force. The body may move directly towards or from 
the centre, or it may move about it in a curved path, called 
the orbit. The latter is the one commonly understood 
when central forces are referred to. Thus, the centre of 
the sun is considered as the centre of the attractive force 
which acts upon the planets, and the planets move in orbits 
about the sun. Similarly, the centre of the earth is con- 
sidered as the centre of the force which acts upon the 
moon, and causes it to move in its orbit. When a stone is 
whirled in a sling, the centre of the force which continu- 
ally pulls upon the body is at the point where the string 
is held by the hand. 

[In the case of the sun and planets, the centre about which 
they revolve is not the centre of the sun, but a point which is at 
the common centre of the system, and which is relatively near 
the centre of the sun.] 

308. Centripetal Force is a name given to that central 
force which acts towards the centre. It is either attract- 
ive, or of the nature of a pull. Thus, the attractive force 
of the sun upon the planets, and of the primary planets 
upon their secondaries, is centripetal. The pull of the 
string upon the body whirled in a sling is centripetal. 
The cohesive force of the metal in a fly-wheel, which 



[309,310.] 



CENTRAL FORCES. 



221 



prevents the rim from flying away from the centre, is of 
the same nature. 

309. When a body moves in an orbit, the centripetal 
force constantly draws the body away from the straight 
line in which it tends to move. According to the first 
law of motion, the body would move in a straight line if 
it were not acted upon by any force, but by the constant 
action of the central force, it moves in a curved line. The 
straight line in which the body tends to move, is a tangent 
to the orbit, as J)G, Fig. 143. 

310. The Orbit. — If a body at rest be left to the action 
of a central force, it will move directly towards or from 
that centre. The same will be true if it be projected 
directly towards or from the centre. In such cases the 
orbit is a straight line passing through the centre of the 
force. 

Thus, if there were a hole through the earth so that a 
body could move from surface to surface, a body placed in 
the hole, or thrown direct- 
ly towards or from the 
centre of the earth, would 
move under the action of 
a central force, and the 
path would be a straight 
line. 

But if the body be pro- 
jected at an angle with 
the line joining the body 
and the centre of force, 

it will move in a curved path. The orbits of ah the 
planets' are ellipses, with the sun at one of the foci. 
Thus, if ADB be an ellipse, representing an orbit of a 




Fig. 143. 



222 KINETICS. [311,312.] 

planet, C and C the foci, then will the sun be at one of 
the latter points in reference to the orbit. It is generally 
assumed that comets move in parabolic orbits, the sun be- 
ing at the focus. This is done so as to simplify the com- 
putations for determining their positions. 

311. To find the orbit it is necessary to know the law 
of action of the central force, and the position, velocity, 
and direction of motion of the body at some point in the 
orbit. In the solar system the force varies inversely aa 
the square of the distance from the centre. See Article 
65. The complete investigation of this subject properly 
belongs to higher mathematics, and we shall consider, in 
this chapter, only that case in which the orbit is a circle. 

312. Centrifugal Force. — This is a name given to a 
force which is equal and opposite to that which deflects 
the body from a tangent to the curve. If the body moves 
in the arc of a circle it will be equal and opposite to the 
centripetal force. If the orbit is not a circle the central 
force will act at an oblique angle to the tangent at nearly 
every point of the path. In Fig. 143, let the body be 
at D, and the centre of the force at O. Let DF repre- 
sent the magnitude of the force, and let it be resolved 
into two forces, one, FF, parallel to the tangent, the 
other, DF, perpendicular to the tangent. If the body is 
moving in the direction DG, the component FF will re- 
tard the motion ; if in the opposite direction, the motion 
will be accelerated ; but in either case the component DF 
pulls the body directly away from the tangent. Then will 
the centrifugal force be equal and opposite to the normal 
component DF. 

[The idea is often conveyed that the centrifugal and deflecting 
forces act upon the moving body at the same time, but such is 



313.J MOTION IN A CIRCLE. 223 

not the case ; for, if they did, they would exactly neutralize each 
other, and the body would move in a straight line. Thus, when 
a train of cars runs around a curve, the forces under which the 
body moves are the propelling- force of the steam and the deflect- 
ing force of the track. The deflecting force is centripetal. The 
so-called centrifugal force is exactly equal and opposite, and is a 
measure, in the contrary direction, of the pressure exerted by the 
rails of the track. If we consider the deflecting force as acting 
betwem the rail and wheel, then will the action of the force upon 
the wheel be centripetal, and the reaction, or pressure outward 
against the rail, be centrifugal. Some have also stated that the 
centrifugal force is that which causes the body to fly away, or 
to tend to fly away, from the centre, as if there were an active 
force radially outward, due to the revolution. But there really 
is no such force in this case. The body tends to go in a straight 
line, tangent to its path. Cut the string of a sling, or let go of 
one string, and the body starts off in a straight line. The term 
centrifugal is not so objectionable as the idea which it has been 
made to represent.] 



Motion in the Circumference of a Circle. 

313. To find the value of the central force when 
the body describes the arc of a circle. Conceive that the 
body is held by a string fastened to a fixed point at the 
centre of the circle ; it is proposed to find the tension of 
the string as the body moves along the circumference. 
To solve this problem, first consider the conditions by 
which a body may be made to move over the successive 
sides of an inscribed regular polygon with a uniform 
velocity. Let ABCD, etc., be a regular inscribed poly- 
gon. If a particle moves over the side AB with a uni- 
form velocity, it is required to determine the direction 
and magnitude of an impulse applied at B, that shall 
cause the body to move along the side BC with the same 
velocity that it did along AB. 

Since the velocity is to be uniform, the sides AB and 



224 



KINETICS. 



[313.] 



BO will be proportional to the velocities along those 
sides ; and the velocity along the chord B O will be the 

resultant of that along AB, and 
of that produced by the re- 
quired impulse. On AB and 
BC, as sides, construct the 
rhombus ABOF; then will 
the diagonal BF be propor- 
tional to the velocity which 
must be produced by the re- 
quired impulse, and will also 
represent the direction in which 
the impulse must act. See Ar- 
ticles 14 and 15. 




Fig. 144. 



Let 



I = AB = BO = the length of a side of the poly- 
gon; 
8 = BF; r = OB = the radius of the circle ; 
t = the time of moving over a side of the polygon ; 
v = the velocity with which the particle moves ; 
m = the mass of the particle ; 
Q = the value of the impulse ; and 
v 1 = the velocity imparted by the impulse. 
The diagonal BF bisects the angle ABO and passes 
through the centre of the circle. Prolong it to G, and 
draw the chords GO and AG. The similar right angled 
triangles BOG and BEG give 

BE __BO 
BG ; 



or, 



BO 

If 
I 



l_ 

2r ; 



1313.] CIRCULAR MOTION. 225 

„\ s = - . 
r 

Multiply both sides of this equation by m and divide by 
£*, and it may be written 

8 P 

m t ? 

_ = m- : 
£ r * 



which gives 



—± — m -. . . . (1) 
t r x ' 



According to Article 122, the first member of this equa 
tion is the force which, acting during a time t (the time of 
moving over one side of the polygon), will produce the re- 
quired velocity v x ; that is, we have 

F=^. . . (2) 

But, in order that the motion shall be along the straight 
line BC, the velocity v 1 must be produced in an instant. 
Let F' be the force which will produce the velocity v x in 
an element of time At ; then, according to Articles 122 
and 124, we have 

^ = ^ = 1. ... (3) 

At At 

If this force acts upon the particle at all the angles of 
the polygon, the particle will describe the successive sides 
of the polygon with a uniform velocity. 

The second member of equation (1) is independent of 
the number of sides of the polygon, and, hence, for the 
same circle and a constant velocity, the ratio of v x to t in 
the first member will remain constant. If the number of 



220 KINETICS. [313.1 

sides of the polygon be increased, the velocity remaining 
constant, the time of describing a side will be diminished, 
hence, v t , equations (2) and (3), will decrease, and F\ in 
equation (3), will also decrease. Let the number cf sides 
of the polygon be increased until the element of time Al y 
during which F' acts, is the same as that occupied by the 
particle in describing one of the sides of the polygon, and 
let Ai\ be the velocity which it will produce in that time; 
then will the value of F' be 

nr Al\ 

F' — m =± ; 
At 

and since At, Ai\, and v are simultaneous quantities, equa- 
tion (1) becomes 

rr/ Ai\ v 2 , A . 

F' = m ^r- 1 = m - . . . (4) 

At r 

Under these conditions the particle describes the sides 
of a polygon of an indefinitely large number oL : sides with 
a uniform velocity. But the limit of the polygons is the 

circle: the limit of the fraction -=J is, according to equa- 

At 

tion (4), the constant quantity — ; and the limit of F' J9 

some constant, whose value equals the limiting value of 
the second member of equation (4). But what is con- 
stantly true of a value as it approaches a limit iihlrjhi- 
itely, is true of the limit ; hence, calling f the limiting 
value of F' ', we have, for thp value of f when the motion 
is in the arc of a circle, 

v 2 
/sin-; . . . (5) 

which is the constant force which acts towards the centre 



[814^B16.j CENTRIFUGAL FORCE. 227 

It follows from this equation that the deflecting force 
varies directly as the mass of the body and the square of 
its velocity ^ and inversely as the rod in* of the circle. 

314. To find the value of the central force in terms of 
the Angular Velocity. 

Let (o = the angular velocity, then 
V = rco ; 
and equation (5) becomes 

f=z mrco 2 . 

315. To find the value of the Cent ri -petal Force in terms 
of the time of a complete revolution. 

The time of a complete revolution is called the Periodic 
time of the motion. 

Let T = the periodic time ; then 
vT=2irri 



irr 



which, substituted in equation (5) of Article 313, gives 

■±7rV 



J == m -Trnr- 



316. Centrifugal Force of Bodies of finite size. — 
Let ?/?!, w?3, etc., be the masses of the particles of the body, 
r u a, etc., be their respective distances from the axis of 
revolution, M the total mass of the body, and ? the point 
at which, if the whole mass of the body were concentrated, 
the centrifugal force would remain the same ; then we 
have 

Mrco = (//?.,/'! + m*r. 2 + etc.)a) = coXmr\ 

Smr 

'• r = -JT> 



228 KINETICS. [317,318.] 

hence, according to Article 216, the centrifugal force will 
he the same as if the whole mass of the body were concent 
trated at its centre of gravity and revolved with the same 
angular velocity. 

Problems. 

317. Find the centrifugal force on the equator due to 

the revolution of the earth on its axis. 

The time of the revolution of the earth on its axis is 
T = 86,164 seconds of mean solar time ;* the equatorial 
radius of the earth is 20,923,161 feet ;+ and w = 3.14159. 
These values in the equation of Article 315 give 

/= 0.1112m. 

The force of gravity exceeds this value ; for it is found 
that a body on the equator weighs (see page 32), 

W=mg = 32.0902m; 

hence, if the earth did not rotate, and other things re- 
mained as at present, a body would weigh 

W, = W+f= 32.2014m. 
We also have 

f_ 0.1112m _1_ 

W x ~ 32.2014m ~ 289 nearlj; 

hence, the rotation of the earth causes a diminution of the 
weight of bodies ; the diminution on the equato 1 * being ^-g 
of their original weight. 

318. In what time must the earth revolve so that bodies 
on the equator will weigh nothing f 

* The earth makes a complete revolution in less than 24 hours mean 
time. 
f See foot note on page 33. 



|319,320.] PROBLEMS. 229 

Let T x be the required time, then will # /J in Article 315, 
equal W l9 and we have 

W x = m j^; 

and dividing the equation of Article 315 by this equation, 
calling T the periodic time of the revolution of the earth, 
we have 

/ _ Tl _ _1_ 

W t ~ T*~ 289 

•■• Ti = ^/A§ T= B T = x L Um - 42 ^ sec - 

319. To find the central (centripetal) force necessary to 
heep the moon in its orbit. 

The mean time of the periodic motion of the moon is 
T = 2,360,585 seconds ; and the mean distance of the 
moon from the centre of the earth is 60.361 R, in which 
R is the mean radius of the earth. Calling B = 20,897,- 
500 feet and substituting these quantities in the equation 
of Article 315, gives, for a unit of mass, 
/= 0.0089. 

320. What is the force of gravity at the moon f 

The force of gravity varies inversely as the square of the 
distance from the centre of the earth. (See Article 65.) 
Calling the mean value of g at the surface of the earth 
82.246, we have 

39 94.fi 

If the data in the two preceding problems were correct 
in every particular, the results should be exactly alike, 
that is, f = g' . In this way Sir Isaac Newton tested the 
law of Universal Gravitation. 



230 



KINETICS. 



[321, 323.; 



321. A locomotive whose weight is W torn runs around 
a horizontal curve whose radius is r feet, with a velocity 
of V miles per hour ; required the centrifugal force y 
that is, the pressure against the o uter rail. 

These quantities, reduced to the units of equation (5), 
Article 313, — that is, to feet, pounds, and seconds — and 
substituted in that equation, give 



/= 



2,000 W 



( 528 ° F v 

\60x60 / 



32i 



W(in tons) x Y\in m.pr. h.) 



= 130 



r (in feet) 



lbs., nearly. 



322. To find the elevation of the outer rail so that 
the resultant pressure of a locomotive, as it passes around 
a curve, will be perpendicular to the plane of the track. 

Let JDE represent the weight, 
W, of the locomotive, and EF the 
centrifugal force ; then will DF 
be the resultant of these forces. 
The elevation of the outer rail 
must be such that BE will be per- 
pendicular to the line joining the 
tops of the rails, which, in prac- 
tice, will be parallel to the line A C. Draw AB horizontal 
and CB vertical. 

Let b == AB-, h = BC; r = the radius of the curve ; 
and v = the velocity of the locomotive. 

Then the similar triangles DEF and ABC give 




or, 



BE:EF::AB:BO; 
W:f::b-h 



PROBLEMS. 



231 



Substitute the value of f, equation (5), Article 313, arid 
W= mg, and we have 



in which v and g are both measured in feet per second, and 
b and r in feet. In practice, the elevation is usually so 
small that b is taken equal to A C, the gauge of the track. 
It will be seen that the elevation is independent of the 
weight of the body, and that it varies as the square of the 
velocity. 

323. To determine the motion of a Conical Pendu- 
lum. — A conical pendulum is a heavy body revolving 
about a vertical axis. The governor of a 
steam engine is an example. The forces 
which act upon the body B are the force 
of gravity, which equals the weight of the 
body, the tension of the piece HA, and 
the force which produces the rotation- 
Let the body be at rest and a force, equal 
to the centrifugal force, act upon the 
body ; then will the relation of the parts 
be the same as before. 

Let W = Bb, and resolve it into two forces, one, Be, 
horizontal ; the other, be, parallel to B A ; then will the 
former represent the centrifugal force, and the latter the 
tension on BA. Let m = the mass of the body and v its 
velocity, then, according to Article 313, 

B G AB sin <£ 




Fig. 146. 



232 




KINETICS. 


We also have 






Bc = 


Wtajk£bo= P^tan^; 




•". m> a r> • — j = ^<7 tan <& ; 
AB sin c/> y ^ ' 


from which 


we find 






VAB .gsmcf) tan </>. 


Let T = 


the time of one revolution, then 




T = 


2tt.BO 

V 






2tt. AB.sm<j> 






VAB . g sin c/> tan <£ 




: 2tt \J — — cos <p ; 


from which 


we find 





T328.] 



hence, the angle is independent of the mass of the body. 

In the governor the balls are not permitted to move out 
freely, but are required to overcome a resistance. The 
resistance may be reduced to an equivalent horizontal 
force applied at B. Let F be this force ; then will 

/= IF tan c/> + F; 

v 2 
.-. F= m -g-p, — IT tan <£ 



= w ( a t • j. - tan H 

\g . AB. sin <j> r J 



[S2S.] PEOBLEMS. 233 

EXAMPLES. 

1. A body revolves about a point in a horizontal plane to 
which it is attached by means of a cord ; required 
the velocity of the body so that the tension on the 
cord shall be twice the weight of the body. • 

2 In the preceding example, if the radius of the circle is 
2 yards, what must be the number of revolutions per 
minute so that the tension on the string will be three 
times the weight of the body ? 

S A body is attached to a point by means of a cord 2 feet 
long, and revolves uniformly in a vertical circle ; re- 
quired the number of revolutions per minute so that 
the tension of the cord shall be zero when the body 
is at the highest point of the circle. 

4. In the preceding example, what will be the tension of 

the cord when the body is at the lowest point of the 
circle ? 

5. A body is placed on the inside of the rim of a wheel 

which revolves about a vertical axis ; if the weight 
of the body is W, the radius of the wheel B, and the 
coefficient of friction fi, what must be the number of 
revolutions of the wheel per minute so that the fric- 
tion due to the centrifugal force will just prevent 
the body from sliding vertically downward ? 

6. A body is in a radial groove in a horizontal disc, and 

attached to the centre of the disc by means of a 
string 30 inches long. The disc is revolved about a 
vertical axis through its centre at the rate of 250 
revolutions per minute ; the coefficient of friction 
between the body and disc being 0.15, what will be 
the tension of the string, the groove being radial. 



234 KINETICS. [328.J 

7. A car runs around a curve whoso radius is 2,500 foot; 

if the rails are in a horizontal plane, and a body rests 
on the floor of the. car between which and the body 
the coefficient of friction is 0.10, what must be tho 
velocity oi' the oar in miles per hour so that the body 
will be just on the point of sliding outward ? 

8. What must bo the elevation of tho outer rail so that a 

car, moving on a curve whose radius is 3,000 feet, 
with a velocity oi~ 80 miles per hour, shall press 
equally upon both rails, the distance between the rails 
being 4 feet 8 inches. 

9. A weight is suspended from a point in the roof of a 

ear by moans of a siring 6 loot long; the oar runs 
around a curve, whose radius is 4,000 loot, at tho 
rate of 40 miles per hour ; how much will the string 
deviate from a vertical on account of the deflecting 
force I 

10. In Fig. 14t>, if AB is 15 inches, and the body makes 

100 revolutions per minute, find the angle /> A 6 y , 
and the distances A .6" and BO. 

11. In Fig. 140, if the resistance which the centrifugal 

force has to overcome is 4 lbs., acting horizontally 
when reduced to the point Z?, the weight oi Z?5 lbs., 
and tho length AJ5, 14 inches; what must be the 
number of revolutions per minute so that the resist- 
ance will be overcome when the angle BAO is 
10 degrees? 

12. If a grindstone whose diameter is 4 foot, thickness 4 

inches, tenacity 000 pounds per square inch, revolves 
about an axis through its centre, how many revolu- 
tions must it make per minute to produce rupture 
along a diameter, no allowance being made foi 
the eye. 



[823.J EXERCISES. 235 



EXERCISES, 

1. If a body moves in a curved line which is not the axe of a circle 

under the action of a central force, will the deviating force be the 
same as the central force ? 

2. Why will the water iu a rotating vessel be highest around the out- 

side of the vessel ? Will this be true of anything besides water ; 
:us grain, or pebbles ? 

3. If the rotation oi' the earth were to oease, about how much would 

the water iu the ocean be raised at tbe poles, and how much would 
it. be depressed at the equator, the earth being oonsidered as fluid? 

4. Does the centrifugal force have any effect upon bodies or particles 

below the surface of the earth, as in a deep mine, for instance ? 

5. If the earth were a hollow sphere and water were thrown into the 

hollow, where would it tome io rest ? 

6. If the earth were to revolve on its axis once in 84 minutes, what 

would happen to bodies on the equator? Would the cohesion of 
the parts prevent, their being thrown off ? 

7. If the moon retained a circular orbit, but should revolve around the 

earth even 15 days, would it. be nearer or more remote from bhe 
earth th:m at present ? 
S. Why do those planets near the sun go around it in less time that 
those more remote ? 

9. Does elevating the outer rail destroy the centrifugal force of a mov- 

ing train ? 

10. Water is put on the face of a grindstone and the stone revolved so 

rapidly that, the water (lies off ; dues it go off radially or tan- 
gent iallv ? 

J 1. Do the centripetal or centrifugal forces have anything to do with 
the velocity with which the stone leaves the sling? 

L2, Clothes may be partially dried by placing them within a perforated 
vessel which is made to revolve very rapidly ; explain the principle, 

13, Are bodies at the poles of the earth affected in their weight on ac- 
count of the rotation ol' the earth ? Show why they weigh more 
there than they would if bhe earth ceased to rotate? 

t-L If a train of cars runs around a circular track in which both rails 
are in the same horizontal plane, is there any danger of the cars 
being overturned by the centrifugal force ? 



CHAPTER XVII. 

SOLUTION OF PROBLEMS IN WHICH THE INTENSITY OF THE 
FORCE VARIES DIRECTLY AS THE DISTANCE FROM THE 



324. When a perfectly elastic solid is pulled or com- 
pressed, or distorted in any manner, the force which resists 
distortion varies directly as the amount of distortion, if the 
elastic limits are not exceeded. See Article 131. This 
law, in which the force varies directly as the distance, 
holds good in several other problems. 



General Formulas. 

325. To find the Velocity. — If a body starts from rest 
and is acted tipon by a force which varies directly as the 
distance from the centre of the force / required the velo- 
city of the body when it reaches the centre of the force, 
and also the velocity at any point of the path. 

Let A, Fig. 147, be the origin of the force, B the point 
where the particle starts, and the 
line AB the path along which the 
particle moves. Erect B G to re- 
present the intensity of the force 
at B, and draw the straight line 
AC; then will any ordinate bo 
represent the force acting upon 
the particle when it is at b. The body, starting from rest 
at B, will be constantly accelerated until it arrives at J, 




1 325.] 



GENERAL FORMULAS. 



237 



but the rate of acceleration will continually decrease from 
B to A, because the force decreases in intensity. 

To represent the velocity approximately by a geometri- 
cal construction, divide the space AB, Fig. 148, into equal 
small spaces, Bf fg, gh, etc., and erect ordinates, fa, gc, 
he, etc. Consider the force as constant while the particle 
is passing from B tof, and represented by the arithmeti- 
cal mean between BO and fa. This force will produce a 
certain velocity, which represent by Bm. Draw run par- 
allel to AB, and at n, where it intersects af prolonged, 




•*V 




Fig. 148. 

draw no to represent the velocity produced by i(af -f eg), 
and draw op parallel to AB, to meet eg prolonged at jp, 
and so on. The points B, n, jp, etc., will be the vertices 
of a polygon ; and by reducing the spaces Bf fg, etc., in- 
definitely, the polygon will approach a curve, BD, as a 
limit. Any ordinate to this curve, as bd, Fig. 149, will 
represent the velocity of the particle when it has reached 
the point b of its path. 

To find a formula for the velocity, let 

vf = the intensity of the force at a unit's distance 
from the centre of 'the force ; 

s 9 = AB = the distance of the particle from A 
when motion begins ; 



KINETICS. £325.] 

s = Ab = any distance from A ; 
v = the velocity of the particle at b ; 
Vi = the velocity of the particle at A ; and 
m = the mass of the particle ; 
then will 

t/So =• BO = the intensity of the force at B; and 
rjs = be = the intensity of the force at b. 
The work done by the force upon the particle while 
moving it from B to A, will be represented, by the trian- 
gle ABO, see Article 96, and hence, will be 

which will impart to the body an amount of kinetic energy 
expressed by 

\mv? ; 

and 

^ 1= Vm*° ; *..'(!) 

hence, the velocity at the centre of the force will vary 
directly as the distance over which the body moves, and 
directly as the square root of the intensity of the force at 
a unit's distance from the centre of the force. 

The work done while moving from b to A will be 

hence, the work done in passing from B to b will be tho 
difference of these, or, 

bo(s 2 - s 2 ) = W ; 

(2) 



V nm, 



2 
"•0 

m 



which gives the velocity at any point of the path. 



[326.] GENERAL FORMULAS. 239 

The last equation may be written 

mv 2 + tjs 2 = w s 2 ; . . . (3) 

and since v and s are variables, and all the other quanti- 
ties constants, it is the equation of an ellipse. Hence, the 
curve BDE, Fig. 149, is an ellipse, of which the semi 
axis AB is s , an d that of AD is found by making s = 0, 
and finding the value of v. This value of v becomes v ly 
and is 



AD 



m 



as given in equation (1). 

After the particle arrives at A it will, by virtue of the 
kinetic energy of the body, pass that point and move on 
until the force at A overcomes the energy, when it will 
stop and return. The distance AE will equal AB. The 
entire distance BE is called the amplitude. This motion 
is called oscillatory or vibratory. 

326. To find the time of the movement of a particle 
from any distance to the centre of the force, when the 
force varies directly as the distance from the centre. 

Equation (2) of the preceding article may be written 

o 9^9 

s 2 — s~ — - V" ; 
V 

which may be represented by a right-angled triangle, in 

which s is the hypothenuse, s one side, and y — v, the other 

V 

6ide. If with A as a centre, and a radius AD = s , a 

quadrant be described, then will BD, the sine of the arc 

CD to the radius of the arc, and AB, the cosine of the 



240 KINETICS. [326.] 

same arc, constantly represent the relation between the 

/ 771 

Bpace s, and the y — times the velocity v. 
V 
The velocity is constantly varying ; hence, at any instant, 
we have, according to Article 10, 



At 



As 



(1) 



in which ^ s is the increment of space passed over in the 
corresponding increment of time. In Fig. 151, let Ah 

= s , BD = s, and AB = )/™v. 

V 

On the line DB, take JDb to represent As, and draw ba 

parallel to AB to meet the tangent Da drawn through D. 




Fig. 150. 




The triangles Dba and ABI) are similar, having the sides 
of the one perpendicular respectively to the sides of 4 he 
other. Hence, we have 



AB\AD\\Db\Da\ 



or 



sFi 



~v:s ::As: Da ; 



GENERAL FORMULAS. 241 



m = \/*L 8 A; ■ (2) 

r m v 7 



which, combined with equation (1), gives 

hence, 

— _ Da 

If Da be diminished indefinitely, it will approach the 
arc as a limit ; hence, an element of the time equals an 

element of the arc divided by \ — s ; and the time of 

v m 

passing from C to A will equal the quadrant ODE, 

divided by y — s v -^ et ^ ^e ^ e required time, then 
m _ 



V- 2 

V mn 



m ° 

Hence, this remarkable result, that tfA<? time of the move- 
ment of a particle from any point to the centre of the force, 
when the force varies directly as the distance from the 
centre, is independent of the distance. The times, there- 
fore, are isochronous. 

Hence, also, the time will be the same for all distances 
from the centre ; and for the same body it will vary in- 
versely as the square root of the intensity of the force at 
a unifs distance from the centre of the force. 

Let t = the time of moving through the amplitude; 
then /— 

t = Zt 1 =irlHl. . . (5) 



M.2 



KINETICS. 



[B21.] 



Problems. 

327. Simple Pendulum.— J. simple pendulum is a 
material particle, suspended by a mathematical line, and 
swinging under the action of gravity. 

Let be the point of suspension, P the position of the 
particle, and w = mg the weight of the particle. Resolve 
the weight w into two components, one, ba, parallel to 

OP ; the other, Pb, perpendicu- 
lar to OP, which will also be tan- 
gent to the arc BA. The line 
OA being vertical, the angle 
baP will equal that at O. The 
component ab will be resisted by 
the tension of the cord OP, and 
the motion will be produced by 
the component bP. We have 
bP = w sin a, 
= mg sin O, 

AP 

= ™<9 op nearly, 

when the angle is small. As OP is a constant /adius, 
it follows that, for small angles of oscillations, the moving 
force varies directly as the distance AP, of the particle 
from the lowest point A. Hence, the time oi an oscilla- 
tion may be determined from equation (5). Tor this pur* 
pose we must find the value of r\. We have 

AP 



e. 




Fi«. 152. 



bP 



mg 



OP 



mg_ 
OP 



v^Ap-TT 

which, substituted in equation (5), and mai mg OP = l 9 
gives 



PEOBLEMS. 243 

which is the value given in Article 67. In practice a 
3ompound pendulum is always used. 

328. Compound Pendulum — Any body of finite size, 
vibrating under the action of gravity, is a compound pen- 
dulum. It is shown, both by analysis and by experiment, 
that there are always two points in a compound pen- 
dulum, about which the body will oscillate in the 
same time, and the distance between these points is 
the length of an equivalent simple pendulum. 
Thus, if the body be suspended at A, and the num- 
ber of vibrations be noted, then there is another 
point, B, at which, if it be suspended, it will vibrate 
in the same time, and the distance AB will be the Fio7i53. 
length of the equivalent simple pendulum. The 
point A is called the centre of suspension and B the cen- 
tre of "oscillation. It requires very accurate measurements, 
and very close observations to determine the length of the 
pendulum which oscillates in a certain time, but it has been 
found many times in different places and the results have 
been recorded. Having found this result, the length of a 
pendulum which will oscillate once in a second may be 
found as follows: 

Let l s =■ the length of the pendulum which will oscillate 
once in a second; then, according to the equation of the 
preceding article, we have 

1 sec. = ir v / ±. 
V 9 

Dividing this equation by the former one, and solving 
for l n gives 



244 



KINETIOi 



[329, 330. J 



From the preceding equation we have 

9 = 11*1*9 

by means of which the acceleration due to gravity may be 
found. 
329. Length of the Second's Pendulum. 

Table giving the Length of the Second's Pendulum at 
different places on the earth, and the acceler- 
ATION due to Gravity at those Places. 









Length of 


Accelerating 


Observer. 


Place. 


Latitude . 


seconds 
pendulum 
in inches. 


force of gra- 
vity ; feet 
per second. 


Sabine 


Spitzbergen 

Hammerfest 
Stockholm 


N. 79°50' 
70°40' 

59°21' 


39-21469 
39 19475 
39-16541 


32 2528 


Sabine 


32-2363 


Svanberg 


32-2122 


Bessel 


Konigsberg 

G-reenwich 


54°42' 
51°29' 


39-15072 

39-13983 


32-2002 


Sabine 


32-1912 


Borda, Biot, and 




Sabine 


Paris 


48°50' 
44°5Q' 
40°43' 
20°52' 


39-12851 
39-11296 
39-10120 
39-04690 


32-1819 


Biot 


Bordeaux 

New York 

Sandwich Islands 


32 1691 


Sabine 


32-1594 


Freycinet 


32-1148 


Sabine 


Trinidad 

Rawak 


10°39' 
S. 0° 2' 


'39-01888 
39 01433 


32-0913 


Freycinet 

Sabine and Du- 


32-0880 






perrey 


Ascension 


7°55' 


39-02363 


32 0956 


Freycinet and 










Duperrey 


Isle of France . . . 


20°10' 


39-04684 


32-1151 


B risbane and 










Rumker. 


Paramatta 


33°49' 


39-07452 


32-1375 


Freycinet and 










Duperrey 


Isles Malouines . . 


51°35' 


39-13781 


32 1895 



330. To find the number of seconds lost by a clock 

vihen carried to a given height above the surface of the 
earik 



[830.] PROBLEMS. 245 

Let the clock on the surface of the earth indicate mean 
solar time, then in one day it indicates 86,400 seconds ; 
when taken to a height above the earth, the vibrations of 
the pendulum will be slower, because the force of gravity 
will be less. 

Let J¥= 86,400, JV ± = the number of seconds indicated 
by the clock when at a height h, t = the time of one vibra- 
tion on the surface, t x = the time of one vibration at the 
height h, and r = the radius of the earth. The length of 
the pendulum remaining the same, we have 



But 



hence, 









Ni = 



r+h' 
which, subtracted from JV, gives 

r+h 
JUT 



T 

1+ l 



The quantity N — JV r is called the rate of the clock 
The quantities r and h inns* be of the same denomination. 



246 



KINETICS. 



[331.] 



If the loss in a day be known we may find the height, 
for we find 

331. To find the time in which a body would pass 
through the earth from surface to surface, if it could 
pass freely without resistances, the earth being considered 
as homogeneous. 

If the earth were a homogeneous sphere, the attractive 

force would vary directly as the distance from the centre 

of the earth ; see Article 78. Hence, 

if r be the radius of the earth, we 

have 

• * = ?; 

which, substituted in equation (5), 
Article 326, gives 




v 9 



(1) 



and in equation (1) of Article 325, gives 

v x = Vgr • ■ • • • (2) 

Equation (1), compared with the value of t in Article 
327, shows that the time required for the particle to pass 
through the earth equals the time of one oscillation, 
through a small arc, of a simple pendulum whose length 
equals the radius of the earth. 

Equation (3) of Article 72 gives v =V2gh ; which, com- 
pared with equation (2) above, shows that the velocity at 
the centre of the earth will equal iV2 times the velocity 
acquired by a body falling freely through a distance equal 
to the radius of the earth, under the action of a constant 




[332.] PROBLEMS. 247 

force equal to the force of gravity at the surface of the 
earth. 
332. Vibration of an Elastic Bar.— Let AB be a 

prismatic bar, having a weight P suspended at its lower 
end. If the weight be pushed up or down 
slightly, or be struck, or in any other manner 
be disturbed in a vertical direction, it will oscil- 
late up and down ; but, in the case of a solid 
bar, the oscillation will be small. The longi- 
tudinal vibrations of rubber, coiled springs, 
and the like, may readily be seen. 

Transverse vibrations, which really follow 
the same law, may readily be seen in the case 
of solid bars, as mav be illustrated by a tuning 
fork. 

Suppose that AB is the length of the bar when P is 
suspended at its lower end, and that by some means it is 
elongated to b. When the disturbing force is removed 
the elastic force in the bar will pull the weight up and the 
end will pass the point B and rise to some point as c, in 
which condition the bar will be in a state of compression. 
The weight will then descend to b, and thus produce a 
vertical oscillation. Let a force F elongate the bar from 
B to b, and let Bb = \, then, according to the equation of 
Article 131, we have 

f=^t\; • • CD 

in which E, K, and I are constants, hence, the elongation, \ y 
varies directly as the pulling force, F The force neces- 
sary to produce an elongation equal to unity, w ill be 

F EK ,0\ 

- = „ = _; . . (2) 



248 KINETICS. [832.] 

which, substituted in equation (1 ) of Article 325, making 
8 Q = Xi, gives 

* = •*/—■; . . ( 3 ) 

or, substituting the value of F, gives 
in which, if m = -P-f-<7, we have 

Let \ be the elongation due to P, and we have 

v 1 = \ S /^.. . . (5) 

The value of rj, equation (2), substituted in equation (5) of 
Article 326, gives 

nor i~ft 

=W*; • • . (6) 

hence, the time of an oscillation is independent of the 
amount of elongation produced by the disturbing force K 

EXAMPLES. 

1. What is the length of a pendulum which will vibrate 

twice in a second ? 

2. What is the length of a pendulum which will vibrate 

once in two seconds? 

3. A pendulum whose length is 39.1 inches, vibrates 



[S82.J EXAMPLES. 249 

86,420 times in a day ; how much must it be length- 
ened to vibrate once each second ? 

4. A second's pendulum carried to the top of a mountain 

lost 45.5 seconds in a day ; required the height of 
the mountain. 

5. A pebble is suspended by a fine thread two feet long ; 

required the time of making 5 oscillations. 

6. If the radius of the earth be 20,923,161 feet, and g at 

the surface is 32.0902 feet, what would be the velo 
city of a body as it passes the centre of the earth if 
it could pass freely through it % 

7. In the preceding example, what would be the time of 

passing from surface to surface ? 

8. A prismatic bar, whose cross-section is J of a square 

inch, length 5 feet, coefficient of elasticity 28,000,000 
lbs., has two weights suspended at its lower ends, 
one of 1,000 lbs. and the other of 3,000 lbs. The 
latter weight suddenly drops off ; required the max- 
imum velocity imparted to the remaining weight by 
the elastic action of the bar. 

9. Required the time of one vibration in the preceding 

example. 

[In these problems the mass of the bar is neglected.] 



CHAPTER XVIII. 

GENERAL PROPERTIES OF FLUIDS. 

333. A fluid is a substance in which its particles are 
free to move among themselves ; as air, water, alcohol, etc. 

A perfect fluid is a substance in which the particles are 
perfectly free to move among themselves, there being no 
friction nor cohesion between them, and in which the least 
force will move any particle in reference to surrounding 
particles. No such substance is known to exist. Even in 
the most volatile gas its particles are supposed to offer 
some resistance between themselves. But the hypothesis 
of perfect fluidity leads to results which are useful in de- 
termining certain formulas applicable to imperfect fluids. 

An imperfect or viscous fluid is one in which there is a 
resistance between its particles. There are all grades of 
viscosity, from that of the most volatile gas to that of solid 
bodies. Mons. Tresca, a French physicist, proved that 
even certain solids, as steel and iron, were somewhat vis- 
cous. If steel be subjected to an immense pressure by a 
blunt tool, the metal, in the immediate vicinity of the 
place pressed, appears to flow like thick tar, or molasses, 
when either is pressed at a point on the surface. Several 
armor plates, ten or twelve inches thick, which had been 
struck by cannon balls, were at the Centennial Exhibition, 
and were excellent examples of the viscosity of metals. 

There are two classes of fluids: liquids and gaseous 
bodies ; the latter of which includes permanent gases and 
vapors, and are called aeriform bodies. 



|334-336.] DEFINITIONS 251 

334. A liquid is a fluid in which there is a slight cohe- 
sion between its particles. Water is taken as a type of 
liquids. The Greek word for water is "vScop, hence the 
science of the statical equilibrium of fluids is called Hydro- 
statics, and of their motion, Hydro dynamics. 

A perfect liquid is a perfect fluid in which there is no 
cohesion nor repulsion between its particles. The hypo- 
thesis of perfect fluidity is assumed unless otherwise stated. 

335. Aeriform bodies are those in which the particles 
mutually repel each other. 

If a vessel, made of elastic material, or provided with a 
piston, be filled with a gas and then enlarged, the gas will 
constantly fill the vessel. There is no known limit to the 
expansion of a gas. The space which it occupies depends 
upon the pressure to which it is subjected. 

336. Forces in the Three States of Matter.— The 

particles which constitute a body act upon each other by 
forces of attraction and repulsion. Supposing that both 
forces exist at the same time, the three states of matter — 
solid, liquid, and gaseous — may be defined by the relations 
which these forces bear to each other. Thus, a solid is a 
body in which the force of attraction greatly exceeds that 
of repulsion ; a liquid, one in which the force of attrac- 
tion equals that of repulsion ; and a gaseous body, one in 
which the repulsion constantly exceeds the attraction. 

Many solids may be reduced to liquids by means of heat ; 
the amount of heat depending upon the degree of attrac- 
tion existing between the particles. Thus, ice, lead, zinc, 
iron, etc., are examples ; and in some cases a substance 
may be made to assume the three states, of which carbonic 
acid is a well-known type. The repulsive f crce may be 
considered as the effect of heat. 



252 FLUIDS. [337-340.J 

337. Law of Equal Pressures. — Thepresiure at any 
point of a perfect fluid at rest is equal in all directions. 

If it were not equal in every direction the particle at 
that point would move in the direction of the resultant of 
the forces. 

338. Normal Pressure. — The pressure of a perfect 
fluid at rest upon the surface of a vessel which contains 
it is normal to the surface. 

For, if it were not, it could be resolved into two com- 
ponents, one of which would be tangential to the surface 
and the other normal to it, and the former would produce 
motion. 

339. Equal Transmission of Pressures. — If a vessel 
contain a perfect fluid at rest, and the fluid be destitute 
of weight, the pressure will be the same at all points within 
the vessel. 

For the pressure against the fluid arises from the reac- 
tion of the sides of the vessel which contains the fluid, and if 
there was a greater pressure at any point above than at any 
other point, motion would result. Gases are so light, that, 
for small quantities under pressure, their weight may gen- 
erally be neglected. Liquids would be without weight if 
subjected to f>e conditions given in Article 77. 

340. i 'h/i -pressure upon the lower part of a vessel which 
contains a heavy fluid is greater than that at the upper 
part. 

A heavy flm d is one that has weight. The force of 
gravity acts do mward on each particle, producing a down- 
ward pressure. This will be equally transmitted to every 
point below it, ut not above it ; hence, the pressure below 
any particular ^article will exceed the pressure above it. 
The downwa*. pressure follows the same general law as 



[341, 342. J 



VERTICAL PRESSURES. 



253 



that of a pile of blocks ; the pressure increases from the 
top to the bottom. 

341. In a heavy, perfect fluid, every pressure, except 
that due to the weight, will be transmitted to every part 
of the vessel without diminution of intensity. 

This may be shown experimentally by means of a closed 
vessel provided with pistons, as in Fig. 156. Let the ves- 
sel be filled with a fluid and the pressure upon the pistons 
be noted. Then, if any piston be pressed inward it will 
be found that, to prevent the other pistons from moving 
outward, the same pressure per square inch must be 
applied to them as to the first piston. In other words, the 
pressure per square inch will be the same on all the pis- 
tons after deducting that due to the weight of the fluid. 





Fig. 158. 

342. Vertical Pressures. — The difference between tits 
pressures on the top and bottom of a vertical prismatic 
vessel, filled with a heavy, perfect fluid, equals the weight 
of the fluid. 

Let the narrow strip ab, Fig. 157, represent a vertical 
prism. If there is a downward pressure on the top at a, 
it will, by the principles of equal transmission, be trans- 
mitted without diminution of intensity to b. The weight 



254 FLUIDS. [343.] 

of the particles will also press downward, and since the 
fluid is supposed to be perfect, the entire weight will be sup- 
ported by the base b. Hence, the pressure at b will equal 
the downward pressure at a plus the weight of the fluid. 

As a result of this proposition, it follows that, if there is 
no pressure at a, the pressure at b will equal the weight of 
the fluid in the prism. If the vessel is filled with a gas it 
must be closed, and there will necessarily be a pressure at 
a, but in the case of a liquid there is not necessarily any 
pressure at that point. • 

343. In Fig. 157, the vertical pressure upon the base at 
c is of the same intensity as that at b ; for the pressure at 
b will be transmitted horizontally to c, but at c it will act 
vertically. The pressure at y will be less than that at c by 
the weight of the fluid in the prism fc ; hence, the pres- 
sure on a portion of the surface at y will equal the weight 
of a prism of the fluid having for its base the area at /J 
and for its height the distance of the point f below the 
top of the fluid, plus the downward pressure upon the 
same area at the top of the vessel. 

The pressure upon the base of a vessel containing a 
heavy , perfect fluid is independent of the form of the ves- 
sel, and equals the weight of a prism of the fluid having 
for its base the base of the vessel, and for its altitude the 
altitude of the vessel, plus a pressure upon each unit of 
the base equal to the pressure per unit upon the upper 
base. 

Let S = the area of the base of the vessel ; 
a = the altitude of the vessel ; 
8 = the weight per unit of volume of the fluid ; 
p = the pressure per unit on the top surface of the 
fluid ; and 



[344,345.] 



RESOLVED PRESSURES. 



255 



P = the total pressure upon the base ; 



then 



P = US+pS. 

344. Resultant pressure against the inside of a ves* 
$e« containing a heavy fluid. 

In Fig. 157, if the points d and e are in the same hori- 
zontal, and directly opposite to each other, the pressure 
upon them will be equal and opposite, and similarly for 
all other pairs of points on the inside of the vessel ; hence 
the resultant pressure upon the whole in- 
terior surface is zero. 

If a hole be cut in one side of the vessel, 
then, as the fluid is being discharged, there 
will be no pressure on that part of the ves- 
sel, and the pressure on the part directly 
opposite will tend to move the vessel in the 
direction of the pressure. If the vessel be 
suspended by a small cord the effect of this 
pressure may be observed. 

It is on this principle that sky-rockets are 
sent into the air. The pressure due to the 
burning of the powder acts upward against the rocket, and 
downward against the air, and the former pressure raises 
the rocket. 

345. Resolved Pressures. — Let the normal pressure 
on APOD be uniform and equal 

to P per unit, and let 6 be the in- 
clination of the surface to the 
vertical ; then will the horizontal 
component of the pressure per 
unit be 

P X = P cos 




Fig. 158. 



B 




^\ * 


/^ 


Vv* 1 


_-J 


If 

Fia. 16a 





256 



PRESSURE OF 



[346, 347 J 



.Let 8 = area ABOD, and S x — the projection of the 
area on a vertical plane passing through A B ; then 

S 1 = S GO80, 

and the normal pressure upon the surface S x will be 
P 1 .S l = Pcos6..Sco&d', 

hence, the horizontal component of the pressure against 
an oblique surface equals the horizontal pressure against 
the vertical projection of the same surface. 

346. Resultant Pressure on a solid body immersed 
in a heavy fluid. — The resultant of the horizontal pres- 
sures will be zero ; for the pressure on one side of the 
body projected in a vertical plane (Article 345) will equal 
that on the other side projected on the same plane. 

If the body be divided into small vertical prisms be, the 
vertical component of the pressure at c will exceed that at 
b by an amount equal to the weight of a 
prism of the fluid whose volume equals 
that of the prism be (Article 342). Hence, 
resultant upward pressure upon the body 
equals the weight of a quantity of the 
liquid of the same volume as that of the 
solid. If the weight of the body be less 
than this pressure, the body will ascend, as is the case with 
a balloon rising in the air, and light wood rising in water ; 
but if the body be heavier than an equal volume of the 
fluid it will fall in it, as in the case of bodies falling in the 
air, or a stone sinking in water. 

347. The point of application of the resultant pressure 
will be at the centre of gravity of the solid, considered as 
a homogeneous fluid. If the solid be not homogeneous, 
its centre of gravity will not coincide with the centre of 
pressure, and if such a body be placed in the fluid so that 




Fig. 160. 



[348.1 



FLUIDS 



257 



its centre of gravity and the centre of pressure are not in 
the same vertical, the body will rotate more or less as it 
rises or falls in the fluid ; for the two forces constitute a 
couple (Article 183), or a couple and a single force (Arti- 
cles 188, 192). If snch a body is not spherical, the com- 
bined motions of rotation and translation will generally 
cause the body to describe a curved path. t 

848. Equilibrium of Fluids of Different Densities. — ■ 
If two fluids which do not mix he placed in two open ves- 
sels which communicate with each other, the heights to 
which they will stand above their common base will be in- 
versely proportional to their densities. 

Let GabB be the vessel, A the common 
base, G the surface of one fluid and B that 
of the other. The portion AD will be in 
equilibrium, hence the pressure of BD will 
equal that of CA. 

Let 8 = the density of the liquid in BD, 
S x = the density of the liquid in CA, 
h = the height BD, and 
\ = the height GA. 

The pressure upon a unit of area of 
the section at A, due to the fluid in BD, will 
be (Articles 342, 343) 

ghh, 
and the pressure due to that in AG will be 

gKKi 

but, there being equilibrium, we have 
gSh = gS^ ; 



Fig 161. 



258 PRESSURE OF [348.J 

This proposition does not apply to the case in which one 
of the fluids is lighter than air, for the vessel containing 
the lighter fluid must be closed at the top. 

EXAMPLES. 

1. A rectangular closed box, whose depth is 1 ft., breadth 

2 ft., and length 3 ft., is filled with a fluid. A cylin- 
drical piston, whose diameter is one inch, is fitted 
into the top of the box ; required the pressure on 
the bottom and sides of the box which would result 
from a pressure of 20 lbs. on the piston. 

2. If a vessel whose base is 6 square inches and height 8 

inches, is filled with water, what will be the pressure 
upon the base, calling the weight of water 62^ lbs. 
per cubic foot % 

3. The lower part of a vessel is a cylinder whose diameter 

is 8 inches and height 6 inches; the upper part is 
also a cylinder whose diameter is 6 inches, and height 
4 inches, the vessel is filled with water and subjected 
to a pressure of 100 lbs. on its upper surface ; required 
the pressure on the base. 

4. A cubic foot of wood that weighs 35 lbs. is placed in a 

vessel of water which weighs 63- lbs. per cubic foot, 
and the body is prevented from rising by a string 
fastened to the bottom of the vessel ; required the 
tension of the string. 
6. In the preceding example, if the body is free to rise, 
and there were no resistance to motion from the fluid, 
what would be its velocity when it has risen 50 feet? 

[The acceleration may be found by Article 86 and the velocity 
by equation (3), Article 24. The result, however, is of no prac- 
tical value, for the resistance of the liquid will be considerable, 
varying nearly as the square of the velocity.] 



[848.] FLUIDS. 259 

6. If a cubic block of stone, whose edges are each 1^ ft 

and weight ISO lbs. per cubic foot, is suspended ir 
water and held by a cord, what will be the tension 
of the cord, the water weighing 62£ lbs. per cubic 
foot '( 

7. One half of a prismatic bar is composed of wood and 

the other half of iron ; the iron being 8 times as 
heavy as the wood, and the whole immersed in a 
liquid whose weight is 65 lbs. per cubic foot, at what 
distance from the middle must a cord be attached so 
that the bar may rest in a horizontal position % 

8. If a prismatic tube is bent as in Fig. 161, and filled 

with mercury to a height DA, how many inches of 
water must be placed in the tube A O to depress the 
mercury three inches, the weight of mercury being 
13 J times that of water. 

EXERCISES. 

1. If a vessel filled with water were placed at rest in a hollow space at 

the centre of the earth (see Article 77), and the vessel should sud- 
denly vanish, would the liquid disperse ? would it remain in the 
same form as that of the vessel before it vanished ? 

2. In the preceding- exercise, if the vessel were filled with a gas, what 

would become of it if the vessel should vanish ? 

3. If a pail were filled with tar would the pressure on the bottom of the 

pail equal the weight of the tar ? 

4. How much less will a heavy body weigh in air than it will :'n a 

vacuum ? 

5 . What must be the weight of a body so that it will neither rise noi 

fall in air ? 



CHAPTER XIX. 

SPECIFIC GRAVITY. 

349. Definitions. — The specific gravity of a body is the 
ratio of the weight of the body to the weight of an equal 
volume of some other body taken as a standard. The 
specific gravity of the standard is taken as unity. 

Distilled water is generally taken as the standard of 
comparison for solids and liquids, and atmospheric air for 
aeriform bodies, but both of these substances change their 
volume for every change of temperature and of pressure. 
It is necessary to fix a standard temperature and pressure. 
Some writers have assumed 60° F. for the standard tem- 
perature for water, while others have taken it at 38.75° F., 
assuming that water at that temperature has its maxi- 
mum density. We will assume the latter as the tempera- 
ture for the standard, although the exact temperature cor- 
responding to the maximum density of water is not 
positively known, it being fixed by some at 38.85° F., and 
by others at 39.101° F. The pressure of the air is deter- 
mined by means of a barometer, and, at the level of the 
sea, equals that of a column of mercury about 29.92 inches 
high. When the pressure and temperature are known, the 
specific gravity may be reduced to the standard. 

The specific gravity of air at 32° F., with, the barometer 
at 30 inches, is about T ^, water being unity. The exact 
relation being established, all substances, including gases 
and vapors, may be compared directly with water as a 
standard. 



[350,351.] 



SPECIFIC GRAVITY 



261 



[The term density, as used in Mechanics, is not identical with 
that of specific gravity, although the ratio of the specific gravi- 
ties of two bodies is the same as that of their densities. The 
specific gravity of a cubic foot of distilled water is unity, but its 
density is 62| lbs. -j- 32^, see Article 85.] 

350. To find the Specific Gravity of a Body more 
dense than that of Water. — Weigh the body in a vacuum 
and then in the standard water ; let 
w — the former weight and w x the lat- 
ter ; then, according to Article 346, the 
weight of a quantity of water equal in 
volume to that of the body will be 




w 



w u 



Fig. 162. 



and, according to the preceding article, the specific gravity 
will be 



w 



w — w x 



absolute weight 
loss of weight ' 



351. To find the Specific Gravity of a Body less 
dense than Water.— Attach it to a body B, which will 
cause it to sink in the water, and let 

w = the absolute weight of the given body, 
w 1 = the absolute weight of the body B, 
w 2 = the absolute weight of both bodies, 

Wx — the weight of B in water, and 

w 2 = the weight of the combined bodies in water. 
Then 

w x — w{ = loss of weight of B, 
w 2 — w 2 r = loss of both bodies, and 

(w 2 — w 2 ') — (w>i — w x f ) — loss of weight due to the 
given body, which equals the weight of a mass of water of 



262 FLUIDS. [352,353.) 

the same volume as that of the given body. Substitute in 
this expression 

w 2 = w + w ly 
and it becomes 

w + Wi — w 2 

w 
,\s = 



w-\-w{— w 2 ' 

352. To find the Specific Gravity of a Liquid.— 

Weigh a body in a vacuum, the water, and in the required 
liquid. 

Let w = the weight of the body in a vacuum, 

w 1 = the weight of the same body in water, and 
w 2 = the weight of the same body in the liquid ; 
then 

w — w 1 = the weight of an equal volume of water, 

and 
w — w 2 = the weight of an equal volume of the 
liquid ; 
but the volumes being equal, we have, from the definition, 

w — w 2 
s= -. 

w — w x ' 

If an empty bottle, whose weight is w, weighs, when filled 
with water, w u and, when filled with the liquid, w 2 , we 
have 

w 2 — w 

8= — ; 

w v — w 

which is the same as the preceding formula. 

353. To find the absolute Weight of a Body.— 

Weigh the body in air and in water. 



[864.J SPECIFIC GRAVITY. 

Let w 1 = the weight in air, 

w 2 = the weight in water, and 
w = the required weight in a vacuum ; 
then 

w — w 1 = the weight of a mass of air equal in 

volume to that of the tody, and 
w — w 2 = the weight of an equal volume of water ; 
then, if s be the specific gravity of air compared with 
water as a standard, we have 

w — w x 
s = -, 

w — w 2 

I0i — sw 2 

.'. w = — ^ -. 

1 — 8 

But as s is very small, less than y^-g-, the value of w foi 
most solids will be very nearly equal to w 1 ; hence, for 
most practical purposes, the weight in air may be used in- 
stead of the absolute weight. 

354. Specific Gravity of a Soluble Body. — Find its 
specific gravity in respect to some liquid in which it is not 
soluble, then find the specific gravity of the liquid in 
reference to water. Let 

s = the specific gravity of the liquid in reference 

to water ; 
s x = the specific gravity of the substance in refer- 
ence to the liquid ; and 
8 2 = the specific gravity of the substance in refer- 
ence to water ; 
then 

s 2 = 5 x .s; 

for if the body is s 1 times as heavy as the liquid, and the 
liquid s times as heavy as water, then will the substance 
be s t times s times as heavy as water. 



264 FLUIDS. 1355-357.] 

355. Specific Gravity of the Air.— Weigh a large 
globe which is filled with air. Exhaust the air as com- 
pletely as possible, the degree of exhaustion being deter- 
mined by a barometric column, and weigh again. Weigh 
the same filled with water. Determine the total weight 
of the air originally in the vessel, and divide it by the 
weight of the water. 

Tins explanation is intended to give only a very crude 
idea of how it may be determined. In determining the 
specific gravity accurately there are many details, a de- 
scription of which is not suited to this work. 

Hydrometers, or Areometers. 

356. Instruments for determining the specific gravity of 
fluids, are called Hydrometers, or Areometers. They are 
cf two kinds, one in which the weight is constant and the 
other in which the volume is constant. 

357. Hydrometer of Constant Weight. — When the 
same instrument is placed in liquids of different densities it 

will sink to different depths, because the weight 
of the volumes displaced must constantly equal 
the weight of the instrument. In Fig. 163, let 
CD be a tube of uniform size, B a hollow ball, 
and A a small vessel containing mercury, so as to 
make the instrument stand upright in the fluid. 
To graduate the stem, place sufficient mer- 
cury in the vessel A so as to make the instru- 
ment float to some definite point and mark it 
1.0. Then float the instrument in a liquid whose 
specific gravity is known to be 1.1 and mark the 
point to which it sinks 1.1. Divide the intermediate space 
into 10 equal divisions, and continue the divisions both 



o.i 




[357.J SPECIFIC GRAVITY. • 2(?5 

above and below as far as desired. This method of equal 
divisions is not exactly correct, as may be seen from the 
following formula. Let 

V = the volume of the part immersed in water ; 
v = the volume included between two consecutive 
divisions of the stem ; 
Di = the density of water ; 
D = the density of the liquid ; 
x = the number of divisions between the point 
marked for water, and the surface when it 
floats in the liquid, and 
s = the specific gravity of the liquid. 
Since the weights of the liquids displaced are constant, we 
have 

g I) l V=gD{V-vx) 

V_ 
_JD_ V 

m '' s ~D 1 ~V-vx~ _ 

v 

Let the instrument be immersed in a liquid of known 
specific gravity, say 1.1, and call it s x , and let x = 10 for 
the space observed, and call its value x 10 ; then we find 
from the preceding formula 

V s t 

V Sx — 1 

Substituting this value in the preceding equation gives 

$1 



s = J^± 



%> 



Si 

-#!<)— X 



^-1 

and letting x = 1, 2, 3, etc., the values of the specific grav 



A 



266 FLUIDS. [358.] 

ity corresponding to the successive spaces maybe com- 
puted and marked on the scale ; or, if desired, the specific 
gravities may be assumed and the spaces, x, computed. 

358. Nicholson's Hydrometer. — This is an areometer 
of constant volume. It consists of a hollow brass cylinder 
A, having a small basket, B, at the lower end 
and carrying a small scale pan, E, at the upper 
end. At C is a small vessel of mercury to 
make the instrument float upright. Sufficient 
mercury is placed in the vessel so that with 500 
grains in the pan, E, the instrument will sink 
to a given notch, D. This instrument may be 
used for determining the specific gravity of solids 
or liquids. If the solid is lighter than water, 
fig. 164. ^q vessel at B is inverted so as to force the body 
down into the liquid. 

To find the specific gravity of a solid, place a small 
quantity of it in the pan E and add weights sufficient to 
sink the instrument to D. Then place the substance in 
the basket B, and the additional weights necessary to sink 
the instrument to the same mark will be the weight of an 
equal volume of water. Let 

w = the weight necessary to sink the instrument 
to D in pure water ; 

w x = the weight which must be added to the sub- 
stance to sink the instrument to the same 
point, when the substance is in the pan E\ 

w 2 = the weights in the pan E, when the substance 
is in the basket B, necessary to sink it to 
the same point, and 
s = the specific gravity of the substance. 



|359.] SPECIFIC GRAVITY. 267 

Then 

w — w x = the weight of the substance in air, 

w 2 — w x — the loss of weight of the substance in water ; 

w — w l 



w 2 — Wi 



To find the specific gravity of a liquid. Sink the in- 
strument to the same depth, D, in water and in the liquid, 
and let 



W = the weight of the instrument, 



w = the weight in the pan E when in water, and 
Wi = the weight in the pan when in the liquid ; 
then 

W+w t 



* W+w' 



Problems. 



359. Mechanical Combinations. — To find the weights 
of the constituents in a mechanical composition when the 
specific gravities of the compound and the constituents 
are known. 

This is a general statement of the noted problem solved 
by Archimedes, in which he determined the respective 
amounts of gold and silver in King Hiero's crown. 

Let w, w ly w 2 , be the weights of the compound and con- 
stituents respectively ; 
s, s i7 s 2 , their respective specific gravities; and 
v, v u v 2y their respective volumes. 

In mechanical combinations we have 

w = w x + Wi ; . . . (1) 
V = Vx + v 8 « • • • (2) 



268 FLUIDS. [360 1 

But 

w = gDv ; w 1 = gD{o x \ w 2 = gZ> 2 v 2 ; . (3) 

which, combined with equation (2), gives 
<ur_w 1 '^' 

i?"A A' ' " ' w 

or since their specific gravities are as their densities, 

W Wx w 2 , 

— = — i .... \0) 

which, combined with equation (1), gives 
(s 2 — s)s 1 

(s t — s)s 2 

360. Chemical Combination. — Two fluids whose vol 
umes are v and v x , and specific gravities s and s x respect* 

vvely,on being mixed, contract - t h part of the sum of their 

volumes ; required the specific gravity of the mixture. 

Let s 2 — the specific gravity required, and 

$ = the weight of a unit of volume of water. 

If there were no condensation the volume after mixture 
would be 

v+Vi; 

but} on account of the condensation, it will be 



( 1 -^ + '«* 



[361,362.] SPECIFIC GRAVITY. %69 

The sum of their weights before mixture will equal the 
total weight after mixture, hence, 

(vs 4- vtfJB = (l \(v + v^Sffi ; 

n vs + v x s x 

72—1 V + V : 

361. Tw £Atf preceding problem the specific gravity of 
the mixture being found, required the amount of conden- 
sation. 

Solving for - gives 

1 __ __ VS + VySx 

n~~ (v + v^ 

362. To find the Specific Gravity of a Body lighter 
than Water when weighed in Air. — A body B ± , whose 
density is less tha?i water, weighs h x grains in air, and B 2 
in water weighs b 2 grains, and B x and B 2 connected, weigh 
c grains in water. The specific gravity of air being 
0.0013, required the specific gravity of B x . 

Let v-l and v 2 be the volumes respectively of B 1 a ad B 2i 
s x and s 2 their specific gravities, and 
8 the weight of a unit of volume of water. 
Then 

(*2 - 1H8 = h • 
(s r - 0.0013)^8 = b t ; 

(s 2 — ik&+(*i - iK$ = c. 

From these we find 

_ b x + 0.0013(5 a - c) 
Sl ~~ b, + b 2 -c ' 



270 FLUIDS. J 362.] 

EXAMPLES. 

1. A piece of wood weighs 12 lbs., and when annexed to 

22 lbs. of lead, whose specific gravity is 11, the whole 
weighs 8 lbs. in water ; required the specific gravity 
of the wood. Ans. -J. 

2. Required the specific gravity of a body which weighs 

32 grains in a vacuum and 25 grains in water. 

3. An areometer sinks to a certain depth in a fluid whose 

specific gravity is 0.8, and when loaded with 60 
grains it sinks to the same depth in water ; required 
the weight of the instrument. 

4. A cubic foot of water weighs 62J- lbs. ; required the 

weight of a cubical block of stone whose edges are 
each 5 ft., its specific gravity being 2.3. 

5. If a body sinks f of its volume in distilled water, what 

is its specific gravity ? 

6. A body, whose weight is 40 grains, weighs 35 grains ir 

water and 32 in an acid ; required the specific grav- 
ity of the acid. 

7. Two pieces of metal weigh respectively 5 and 2 lbs., 

and their specific gravities are 7 and 9 ; required 
the specific gravity of the alloy formed by melting 
them together, supposing that there is no condensa- 
tion. Ans. 7.474. 

8. A. compound of gold and silver weighing 10 lbs. has a 

specific gravity of s = 14, that of gold being s t = 19.3, 
and of silver s 2 = 10.5 ; required the weight w 1 of 
the gold and w 2 of the silver. 

Ans. w x = 5.483 lbs., w 2 = 4.517 lbs. 

9. If 73 lbs. of sulphuric acid, the specific gravity of which 

is 1.8485, are mixed with 27 lbs, of water, and the re« 



] SPECIFIC GRAVITY 271 

suiting dilute acid has a specific gravity of 1.6321j 
what will be the amount of condensation ? 

[Before the formula in Article 361 can be used, it will be 
necessary to express the quantities in terms of volumes. Let 
1 lb. of water be a unit of volume, then will the volume of water 
be % = 27, and of the acid v t = 73 -*- 1.8485 = 39.4915 ; and in 
the formula, 8=1.] 

Ana. 0.0785. 

10. A body B 2 weighs 10 grains in water, and B t 14 
grains in air, and B l and B 2 together weigh 7 grains 
in water ; required the specific gravity of B u that 
of air being 0.0013. Ans. 0.8237. 



EXERCISES. 

1. If a body floats at a certain depth in a liquid when the vessel which 

contains it is in the air, will it sink to the same depth when the 
vessel is in a vacuum ? 

2. Why will smoke sometimes rise in the air ? Why will it fall at other 

times ? 

3. Will the depth to which a body floats in a liquid be affected by 

changes in the density of the air ? 

4. If a rubber bag containing a gas be made to just sink in a liquid, 

will a pressure on the surface of the liquid condense the gas ? and 
if so will it have a tendency to rise ? 

5. Considering the compressibility of iron and of water, can iron sink so 

deep in water as to float at that depth ? or, in other words, will 
the water become as dense as the iron ? 

6. If water were incompressible, is there any limit to the depth to 

which a body heavier than water, and also incompressible, will 
sink in the water ? If the body were compressible, is there a limit ? 

7. If an egg will sink in pure water, will it float or sink in brine ? What 

must be the condition of t\\e brine that the egg may float between 
the top and bottom ? 

8. Will a vessel of water which contains a fish weigh any more than if 

the fish were removed ? 



CHAPTER XX. 

HYDROSTATICS. 

363. Compressibility of Liquids. — The mechanical 
properties of liquids are determined on the hypothesis 
that liquids are incompressible. They are, however, more 
compressible than most solids. If a cubic inch of watei 
be pressed with fifteen pounds on each and every side, the 
volume will be diminished ^oioi, hence one pound to the 
square inch will diminish the volume 3-0-0V o-o- If the water 
be confined in a perfectly rigid, prismatic vessel, the com- 
pression would take place entirely in the direction of the 
length, and would equal 3-0 A^o of the length* for every 
pound per unit of area of the end pressure. Water, there 
fore, is nearly 100 times as compressible as steel. See 
Article 130. All other liquids are more or less compres 
sible, yet, for most practical purposes, they may be con 
sidered as non-elastic without involving sensible error 
Liquids are sometimes defined as non-elastic fluids. 

The first experiment, to determine the compressibility 
of water, was made by a philosopher at Florence, Italy. 
He filled a hollow globe made of gold with water, and 
then subjected it to a great pressure, thereby flattening it, 
This diminished the volume, and it was observed that the 
water oozed out through the pores of the gold ; from whicii 
he drew the erroneous conclusion that the liquid was not 
diminished in- volume. 

364. Free Surface. — The upper surface of a liquid 
contained in a vessel which receives no pressure, is called 

* More accurately, totSctt* 



[365.J PROBLEMS. 273 

the free surface. The upper surface of water in the 
atmosphere is pressed downward by the air with about 
fifteen pounds to the square inch ; yet such a surface is 
often considered as a free surface. 

The free surface of small bodies of a perfect liquid at 
rest may be considered as horizontal ; for it will be per- 
pendicular to the direction of action of the force of grav- 
ity, Articles 338 and 204 ; but for large bodies of a liquid 
it is spherical, partaking of the general form of the sur- 
face of the earth. 

365. A Level Surface is one which cuts at right angles 
the resultant of the forces which act upon its particles. 
Thus, in a vessel filled with a heavy liquid at rest, it is 
horizontal ; in the ocean it may be a surface at any depth 
and nearly concentric with the free surface ; in the second 
problem below it is a paraboloid of revolution, etc. 

Problems. 

1. A vessel is filled with a perfect ', homogeneous liquid, 
and drawn horizontally with a uniform acceleration / re- 
quired the form of the free surface. 

Let F be the force producing an acceleration /J and M 
the mass of the liquid. Then, according to Article 86, 
we have, for the horizontal 
force, 

F=Mf=W^=cb; 

and for the vertical force, 

W=ob. 

These forces will be uniformly distributed throughout 
the mass ; hence the resultant of the forces on the parti 




274 



HYDBOSTATICS 



r365.] 



cles will be equal and parallel to each other, and also nor* 
mal to the free surface, Article 338 ; therefore the free 
surface will he a 'plane. The level surfaces will also be 
planes parallel to the free surface. 

Let <f> be the inclination of the free surface to the hori- 
zontal, also == cob ; then 

tang 6 = —£ — I. 

% If a cylindrical vessel containing a perfect \ homoge- 
neous liquid be revolved uniformly about a vertical axis, 
what will be the form of the free surface f 

The vessel may be any solid of revolution, the axis of 
-evolution coinciding with the axis of rotation. Any 

element of the liquid will be acted 
upon by two forces ; one, the force 
of gravity acting vertically down- 
ward and equal to oh, the weight, 
w, of the particle ; the other, the 
centripetal force, acting horizon- 
tally and radially inward. Let the 
vessel be at rest and a force equal 
to the centrifugal force act upon 
the particles. 

Let co be the angular velocity, and r the distance of any 
particle from the axis of rotation ; then, according to 
Article 314, the centrifugal force will be 




Fig. 166. 



w 



oa = mra? = — ra> 2 = be ; 
9 



and the vertical force, 



6b = w. 



[365.] PROBLEMS. 27S 

The resultant of these forces must be normal to the free 
surface. Prolong the line of the resultant co, until it 
meets the axis at D ; then, from the similar triangles obc 
and o OJ? y we have 



or 



DG 
Go 


oh 
~~ cV 


DG__ 
r 


w 

w ,' 
— rar 

9 


.\DG 


_ 9 . 



that is, the subnormal, DG, is constant. It is shown in the 
Calculus that the parabola is the only curve which pos- 
sesses this property ; hence the surface is a paraboloid of 
revolution. All the level surfaces are equal paraboloids. 

It is also shown in the Calculus that the volume of a 
paraboloid of revolution is one-half that of a circumscribed 
cylinder ; hence, if the cylinder be at rest, the free sur- 
face will be midway between the highest and lowest 
points of the paraboloid. 

3. If a perfect, homogeneous mass of liquid he acted 
upon hy a force which varies directly as the distance from 
the centre of the mass, what will he the form of the free 
surface f 

It will be a sphere ; for the force at the surface will 
then be equal and normal at every point of it. 

4 If in the preceding example, the mass rotates ahout 
an axis, what will he the form of the free surface ? 

The force of gravity will act directly towards the cen- 
tre of the mass, and the centrifugal force will act outward, 



276 HYDROSTATICS. [365.] 

perpendicular to the axis of rotation, and the resultant of 
these forces must be normal to the surface. This prob- 
lem is approximately that of the earth, and its solution 
involves higher mathematics. The form is, approximately, 
an ellipsoid of revolution, and is often called an oblate 
spheroid. 



EXAMPLES. 

1. A vessel containing a liquid, whose weight (including 

the liquid) is 50 lbs., is drawn horizontally by an 
effective moving force (Article 87) of 15 lbs. ; re- 
quired the inclination of the surface to the hori- 
zontal. 

2. A rectangular box 3 feet long contains a quantity of 

liquid. If the liquid is one foot deep, what must be 
the acceleration of the box in a horizontal direction 
that the free surface at the forward end shall just 
touch the bottom of the vessel ? 

3. In the preceding example, if the rear end of the box 

slopes outward at an angle of 45 degrees, what must 
be the acceleration of the box so that all the watei 
• shall escape by flowing over that end ? 

4. In Fig. 166, if the vessel is cylindrical and 2 ft. in dia- 

meter, and the free surface of the liquid is 3 inches 
from the top, what must be the number of turns per 
minute so that the upper edge of the surface shall 
just reach the edge of the vessel ? 
5 If the vessel is rotated 30 turns per minute, what will 

be the equation of the parabola ? Q 

Ans. ^ = 2^x. 



.] 



liAW OF PRESSURE. 



277 



Law of Pressure. 

366. The pressure of a perfect, homogeneous liquid 

varies directly as the depth below the free surface. 

Since such a liquid is incompressible, we may consider 
a vertical prism of the liquid as composed of blocks of equal 
size and weight, placed one above the other. The first 
block will press with its entire weight upon the second 
one, and the first and second upon the third, and so on, and 
since the weights are equal to each other, the pressure 
upon the succeeding blocks will vary as the number of the 
blocks, or as 1, 2, 3, etc. 

Draw a horizontal line, la, to represent the pressure at 1, 
then will 25, representing the pressure at 2, be twice as 
long as la ; 3c, three times as long, and so on. In this case 



— f 



— d 



~i e 

• L -v 



%fifr 




Fig. 167. 



Fig. 168. 



.the pressure does not increase continuously, but by steps. 
If now the blocks be divided indefinitely, the steps will 
become indefinitely small, and ultimately may be repre- 
sented by Fig. 168, in which AE represents the pressure 
at A, and any horizontal line drawn from AB to BE, the 
pressure at that point. Thus far we have considered the 
vertical pressure only, but in a perfect liquid the pressure 



278 HYDROSTATICS. [367, 368. J 

will be the same in all directions, Article 337 ; hence the 
truth of the proposition. It follows from this that : 

367. The pressure against an elementary area equals 
the weight of a prism of the liquid whose base is the area 
pressed and whose altitude is the vertical distance of the 
area below the free surface. 

Let Aa = be the area, 

h = the distance below the free surface, and 
8 = the weight of a unit of volume ; 

then the pressure will be 

8. Aa.h. 

368. To find the pressure of a liquid against a ver- 
tical rectangle in which one edge coincides with the free 
surface. 

Let ABGD represent the rectangle, in which the side 
BO coincides with the surface of the liquid. Draw the 
horizontal line AE to represent the 
pressure at A, and draw BE; then 
will the triangle ABE represent 
the pressure against the line AB. 
Complete the triangular wedge 
ABOB-FE; the volume of this 
wedge will represent the entire pres- 
sure against the rectangle. 

Let 6 = the weight of a unit of volume of the liquid ; 
h=zAB; b = AB. 

The pressure on a unit of area at A will be, according tc 
Article 366, 

AE=8h; 
hence, the volume of the wedge will be 




1.869, 370. J LAW OF PRESSURE. MU 

bh.%8h = i8bh 2 . 

If the liquid be water, this expression becomes 

31ibtf lbs. 

in which b and h are in feet. 

We see from this expression that the entire pressure 
from the free surface downward, varies as the square of 
the distance from the surface. 




). To find the pressure of a liquid against a vertical 

rectangle when the upper edge is parallel to the free 
surface. 

Let h^BJST; h=AN; b =AD ; 

then will the pressure on the rect- 
angle ABGD equal the difference 
of the pressures on ANMD and 
VjBJVM, or 

%8b(h? - h*). 

370. Pressure on any Surface. — Conceive the surface, 
whether plane or curved, to be divided into small areas. 

Let 8i, s 2 , <%, etc., be the areas ; 

hi, h 2 , h s , etc., their respective distances below the 

free surface ; 
S = «! + s 2 4- s 3 + etc. = 2s, be the entire area of 

the surface pressed by the liquid ; 
x, the depth of the mean pressure ; and 
8, the weight of a unit of volume. 

Then, according to Article 367, the pressure upon the sur 
face will be 

8s Jh. -b 8s 2 h 2 -f 8s. d h 3 = 8Ssh ; 



280 



HYDROSTATICS 



[370.J 



and this also equals the total area into the mean pressure, 
hence 

SSoC = SUsh ; 

_ 2sh 

hence, according to Article 216, x is the depth of the ceiir 
tre of gravity of the surface. Therefore, the total pressure 
on any surface, 8, equals the weight of a prism of the 
liquid whose base is the area of the surface pressed, and 
whose altitude is the depth of the centre of gravity of the 
surface below the free surface. 

Problems. 

1. Triangular Surfaces. — To find the pressure against 
a triangular surface whose base is parallel to the free 
surface, and whose apex is in that surface. 

Let ABC be the triangle, b — AC, h= BD, and g the 





centre of gravity of the triangle ; then, according to Arti- 
cle 370, we have 

8.ibh.%h=i8bh\ 

The pressure is also represented by the volume of the 
pyramid B—ACFE, or 

hAE.AC.\BD = SLb.ih = i8W, 
as before. 



[370.] PROBLEMS. 281 

2. Let the base coincide with the free surface. TheD 
we have 

8.ibh.ih = \hbh\ 

This may also be represented by a triangular pyramid whose 
base is ABO, Fig. 172, and whose altitude is AE '= 8h. 

3. Cones. — Find the normal pressure upon the concave 
surface of a closed cone filled with a liquid / (1) with t.*e 
axis vertical and apex uppermost : (2) axis vertical ana- 
cone inverted ; (3) with the axis horizontal / (4) pressure 
on the base in case (1) / (5) vertical pressure on the con- 
ca/ve surface in (1) / (6) weight of the liquid. 



Let r — 


the radius of the base, and 




h = 


the altitude of the cone. 




Then 






(1) 


%hirrhVr* + tf. (±) 


hirr^h. 


(2) 


i*irrhVr> + h\ ( 5 ) 


f&r**A. 


(3) 


$irr 2 V?2 + h 2 . ( 6 ) 


iBirr^h. 



Observe that the weight equals (4), the downward pres- 
sure, minus (5) the upward pressure. 

4. Spheres. — A sphere is submerged in a liquid : find 
the normal pressure upon the external surface (1) when it 
is just submerged ; (2) when submerged to any depth / 
(3) weight of a quantity of the liquid equal in volume to 
that of the sphere. 

Let r = the radius of the sphere, and 

h = the depth of the centre of the sphere below tne 
free surface. 
Then 

(1) 4&rc*; (2) 4Bm*h; (3) |07r^. 



282 HYDROSTATICS. [371.J 

EXAMPLES. 

1. In Fig. 170, if the edge MN of the rectangle coincides 

with the surface of the liquid, and AN is 3 feet, 
how far from the surface must the line OB be drawn 
so that the pressure on the two parts shall be equal \ 

Ans. 2.121 feet. 

2. A rectangle whose sides are 1.4 feet and 2.6 feet re- 

spectively, is immersed in water with the former 
side in the surface, and is inclined at an angle of 
56° 35' to the free surface ; required the pressures 
on the parts into which the rectangle is divided by 
its diagonal. 

3. A cylinder whose base is 2 feet in diameter and alti- 

tude 3 feet, is filled with water; required the pres- 
sure on the concave surface, the pressure on the 
base, and the weight of the water. 

4. A sphere 10 feet in diameter is filled with water ; re- 

quired the normal pressure on the interior surface s 
and the weight of the fluid. 

5. Find the pressure on a rectangular submerged flood- 

gate, ABOD, Fig. 170, whose depth, BN, is 10 ft., 
height of the gate, AB, 3 ft., and width, BO, 2 ft. 

6. In the preceding example, find the pressure if the top 

of the floodgate is also submerged on the opposite 
side to a depth of 4 feet. 

Centre of Pressure. 

371. The centre of pressure of any surface immersed 
in a flui4 is the point of application of the resultant of all 




[372-375.] CENTRE OF PRESSURE. 283 

the pressures upon it. It is, therefore, that point in an 
immersed surface to which, if a force equal and opposite 
to the resultant of all the pressures upon it be applied, 
this force will keep the surface in equilibrium. 

372. Rectangles. — Let the surface be a rectangle in 
which one end coincides with the free surface of the liquid ; 
then will the centre of pressure be at 

two-thirds the depth of the rectangle. 
'Ch.3 total pressure may be represented 
by a wedge whose end is the triangle 
ABJ3. Hence the centre of pres- 
sure will be at the same depth as that F _ 
of the centre of gravity of the triangle 
ABE. Let C\ on the vertical line 

' a ig. l to. 

BA, be on a horizontal line through 

the centre of gravity, g, of the triangle ABE. Then, 

according to Article 222, B C will be %BA. 

373. Submerged Rectangle. — In Fig. 170, the centre 
of pressure will be at the depth of the centre of gravity of 
the trapezoid ABGE. Since BG and AE are directly 
proportional to NB and JVA, we have, from Example 6, 
page 146, for the depth required, 

JTA + UTB 

NA~,AB NA + MB . 

374. Triangles. — The centre of pressure of the triangle 
in Fig. 171, will be opposite the centre of gravity of the 
pyramid B— A CFE\ or 

Bg = IBB. 

375. The centre of pressure against the triangle ABO, 
when OB is in the free surface, is at the centre of grav- 
ity g, of the wedge ABO —E. To find this point, we 
pbserve that this wedge is what remains after removing 



284 



HYDROSTATICS. 



[376-377.1 



the pyramid BOGF—F, Fig. 175, from the wedge 
BCGF—AF. The centre of gravity of the large wedge 
is at one-third its altitude, and of the pyramid at one< 





Fig. 174. 



Fig. 175. 



fourth its altitude from the base ; hence, according to Arti- 
cle 224, we find that the centre of pressure, g, is at one-half 
the altitude from the base OB. 

Flotation. 

376. Plane of Flotation. — If a body in a liquid is 
lighter than the liquid, it will float, and the conditions of 
equilibrium will be determined according to Article 346. 
The intersection of the plane of the free surface with the 

floating body is called the plane 
of flotation. The line joining 
the centre of gravity of the 
solid and of the displaced 
liquid is called the axis of 
flotation. 

377. Conditions of Equili- 
brium of a Floating Body. — 

One condition is, according to 
Article 346,that the weight of the displaced liquid shall 
equal that of the body. Another is that the axis of flota- 
tion shall be vertical. 




Fig. 17( 



[378,379.] FLOTATION. 285 

378. Stable Equilibrium.— In Fig. 176, let C be the 
centre of gravity of the displaced liquid, and G that of the 
body when the axis of flotation is vertical; and C the 
centre of gravity of the displaced water when the axis is 
inclined. Let the vertical through C meet the line G V 
ir. the point M. When the body is turned through an in- 
definitely small angle, the point M is called the metacen- 
tre. When M is above G the pressure of the fluid up- 
wards along CM, and of the body downwards along the 
vertical G W, tend to bring the body back to the position 
in which the axis of flotation will be vertical. Hence the 
equilibrium is stable when the metacentre is above the 
centre of gravity of the body. Observing that C G is a 
new axis of flotation, it follows that the equilibrium is 
stable when the axis of flotation turns in a direction oppo- 
site to that of the rotation of the body when the position 
of the body is disturbed. 

379. Depth of Flotation. — Let D be the density of the 
body, V its volume, and s its specific gravity ; and D u V u 
s l9 the corresponding quantities for the displaced liquid. 
Then, according to Articles 85 and 349, and the first con- 
dition of Article 377, we have 

gDV=gD 1 Y 1 ; 

•••>.=f"=i>'- 

If the body floats in pure water then s x = 1 and 
V 1 = sY. 

Problems. 

1. Let the body be a right cone with the axis verttcal 
and apex upward ; required the depth of flotation. 



286 HYDROSTATICS. 1379. 

Let r = the radius of the base, 
h = the altitude, and 
x = the depth of flotation. 
Then 

j (h — x) = the radius of the plane of flotation. 

hence 

and the equation of the preceding article becomes 

/ s 

th S\ 



-M-5>- 



2. A rectangular wall whose height is hfeet, thickness 
bfeet, and weight of a cubic foot of the masonry 8 pounds, 
resists the pressure of water whose height behind the wall 
is h x feet ; will the wall be stable in reference to slipping 
on its base, or to overturning about its outer edge f 

The pressure of the liquid for a unit of width will be, 
according to article 368, 

ix62|A 1 2 ; 
and the centre of pressure, according to Article 372, will 
be at ^h x from the base, hence the moment of pressure in 
reference to the outer edge of the wall will be 

ix20|A 1 8 . 
The weight of the wall will be 



[879] FLOTATION. 287 

and the moment in reference to the lowest outer point 
will be 

hence, the wall will be stable in reference to rotation, if 

SJ 2 A>20|A 1 8 ; 
and in reference to slipping on the base, if 

in which jj, is the coefficient of friction of the wall on its 
foundation ; see Article 107. 

EXAMPLES. 

1. In Fig. 170, what is the depth of the centre of pressure 

below MN] the end JOT coinciding with the free 
surface, and AN being 3 feet? Ans. 2 feet, 

2. In Fig. 170, if ABCD is a floodgate, BN being 5 feet, 

and AB 10 inches, how far below B must a hori- 
zontal bar be placed so as to balance the pressure 
against the gate ? 

3. If the height of a rectangular wall be 8 ft., weight of 

the masonry 180 lbs. per cubic foot, what must be 
the thickness of the wall to resist overturning from 
the pressure of water when level with the top of the 
wall? 

4. A. triangular wall whose base is 4 ft., height 8 ft, 

weighs 120 lbs. per cubic foot ; required the height 
of water which it will sustain in reference to over- 
turning, and leave a coefficient of stability of 2. 

[The degree of stability will be determined by assuming 125 
lbs. per cubic foot for the weight of the water.] 



288 



HYDROSTATICS 



r 379.1 



The space between the pistons .27 and A is filled with 
water, and a pressure of 500 lbs. 
is exerted on the small piston 
by means of a lever ; if the dia- 
meter of the small piston is 1J 
inches, and of the large one 15 
inches, what will be the pressure 
exerted by the large piston ? 

[This machine is called a Hydiaulio 
Press. It was perfected by one Bramah, 
who packed the pistons with a leather 
collar in such a way that the pressure of 

the water forced the leather against the sides of the cylinder, thus 

keeping the pistons water-tight.] 




Fig. 177. 



EXEEOISES. 

1. In Figs. 165 and 166, will the surface be the same for mercury as 

for water, other things being the same ? 

2. Why do the answers to Examples 1, in Articles 370 and 379, differ ? 

3. If a vertical square be entirely submerged, will the centre of pres- 

sure coincide with the centre of gravity of the square ? 

4. In the preceding Exercise, if the square be turned about a horizontal 

line, passing through its centre of gravity, will the depth of tha 
centre of pressure be changed ? 



CHAPTER XXL 

HYDRODYNAMICS. 

380. Mean or Average Velocity. — The velocity is not 
the same at all points in the cross section of a stream, 
whether the stream be a river, or canal, or in a pipe or 
tube. That velocity which, being multiplied by the area 
of the cross section, will equal the quantity discharged is 
called the mean velocity. 

Let Q = the quantity which passes a section, 
S = the area of the cross section, and 
v = the mean velocity ; 
then 

vS= Q; 

381. Permanent Flow. — If the same quantity passes 
all the transverse sections the flow is said to be permanent : 
otherwise it is variable. In a canal the flow would be per- 
manent if there were no wastes from evaporation or leak- 
age ; in a pipe without branches the flow will be perma- 
nent throughout its length. 

382. Variable Velocities. — In a stream of variable 
sections in which the flow is permanent, the mean veloci- 
ties are inversely proportional to the transverse sections of 
the stream. 

Let v, S, and Q be the quantities for one section, 
Vi, /Si, and Q the quantities for another section; 
then, according to Article 380, we have 
13 



290 HYDRODYNAMICS. [ 383. 1 

vS = Q = v&; 
m v__8 } 
'''v ± ~ 8' 

383. Problem. — To find the velocity with which a per- 
fect liqmd will flow through an orifice in the base of a 
vessel. 

We will assume that the particles start from rest at a 
point near the orifice, and that the velocity of their exit 
is produced by a constant pressure, equal to the weight of 
the water vertically over them (Article 367). 

Let ab be the short distance through which the 
velocity is generated, 
AB = A, the height of the liquid, 
W, the weight of the quantity in the height ab, 
8, the area of the orifice, 
a§\- A -F, the pressure of the liquid above the orifice ; 

Fig. 178. and 

v, the velocity of discharge. 
Then, since the flow is supposed to be without resistance, 
the conditions are essentially the same as Problems 1 and \ 
pages 45 and 46 ; hence 

.ab 



ZEg.t 

v =y~w 



But in this problem we have 

F= 88h, 

W=B8.ab; 
which substituted above gives 

v = V2gh, 

which is the same as that of a body falling freely through 
a height h ; see Article 72. 



[884-386.] DYNAMIC HEAD. 291 

384. The velocity through an orifice in the side of a 
vessel will also be 

v=V2gh; 
in which h is the depth of the orifice below the surface ; 
fcr the pressure against the side is the same as the vertical 
pressure at the same depth. 

385. Head due to a Velocity. — The preceding equa« 
tion gives 

in which the height A, corresponding to the velocity v 9 is 
called the head due to the velocity, or simply the head. 
The corresponding velocity is called the velocity due to the 
head. 

386. Vertical Pressure on the top of a Vessel.— 
If a piston were fitted into the vessel at B, Fig. 178, and 
a pressure applied to it, the velocity of issue would be in- 
creased. To find the resulting velocity, let 

P = the pressure on the surface, 

S = the area of the upper surface, 

j) = P + S= the pressure on a unit of area, 

Ax = a height of liquid which will give a pressure 

equal to P, 
D = the density of the liquid, and 
8 = the weight of a unit of volume of the liquid, 

= Dg (see the last equation of Art. 85) ; 
then we have 

P=BSh 1 ; 

and the velocity of flow will be 



v=V2g{h+h t ). 



292 



HYDRODYNAMICS. 



[387-339. 



This is called finding an equivalent head. If the liquid 
issues into a fluid more dense than air, there will be a 
counter-pressure. If A 2 is the head due to the difference 
of the pressures of the air and fluid, then 



v = V2g(A-h 2 ). 

387. Pressure of the Air. — When a vessel is in the ail 
it is pressed on the upper surface with nearly 15 pounds 
to the square inch, which is equivalent to a column of 
water of the same base and about 34 feet high. If a ves- 
sel of any liquid should discharge into a vacuum, this head 
must be added to the head of the liquid, but in practice 
the air presses against the issuing stream with the same 
pressure per unit that it presses against the top, so that the 
head due to the pressure of the air is not considered. 

388. Vertical Jet. — If the issuing jet should be verti- 
cally upward, as in Fig. 179, and there were no resistances 







D 
E 
F 






_. = 4 


Jfcr 




KfH 




t---r^._M 


J> 


I G 


Fig. 


179. 




Fig. 180. 





from the air or the sides of the orifice, the jet ought tc 
rise as high as the free surface of the liquid in the vessel. 
But it is found in practice that it always falls short of 



that height. 



389. Orifices in the Side of a Vessel.— If the fluid 
issues horizontally from an orifice in the side of a vessel, 
the jet will be subjected to the same law as that of a pro- 



[890-1.] COEFFICIENTS OF DISCHARGE. 



293 



jectile thrown horizontally. Let D, Fig. 180, be an orifice 
in the side of a vessel, and DG the path of the fluid vein. 
In Article 306 make x — AG and y = DA, and we have 



AG = %Vh.DA. 

But h = BD, hence 

AG = 2VBD.DA. 

If on AB, as a diameter, a semicircle AHB be de- 
scribed, and an ordinate DI be erected, then, from geome- 
try, we have 

DI=VBD.bA\ 
hence, the range 

AG = 2DI; 

that is, if a semicircle be constructed on AB as a diame- 
ter, the range will be twice the ordinate of the semicircle 
drawn from, the orifice. 

Hence the maximum range AC will result from the 
flow through an orifice, E, at the middle of the height. 

Also, the ranges from two orifices, D and F, equidistant 
from the centre, F, w T ill equal each other. 

390. Oblique Jet. — If the jet be oblique, spouting up- 
ward or downward, the range may be determined by the 
formulas in Article 299, considering the vein as the path 
of a projectile. 

Coefficients of Flow. 

391. Coefficients of Contraction. — If the vein issues 
through a thin plate, the smallest part of the vein will be 
at a short distance from the orifice. It appears that the 
particles, as they approach and issue from the orifice, tend 



294 HYDRODYNAMICS. [392.1 

to cross each other's path, and by thus interfering with 
each other first produce contraction and afterward expan- 
sion, as shown in Fig. 181. 

Let S be the area of the orifice ab, 
s \\\u[c^ $ij the area of the contracted part cd, and 

^ppSgllP 1 m^ the coefiicient of contraction ; 



then 

Fio. 181. 

For a very thin plate, the distance of smallest section 
of the vein from the orifice will be equal to the radius of 
the orifice, and the diameter of the smallest section will 
be 0.8 of the diameter of the orifice, and the coefficient of 
contraction will be the square of 0.8 ; hence for a thin 
plate m 1 = 0.64. 

For an adjutage, that is, for a short tube, aodc, whose 

length is two or three times the diameter of the orifice, 

attached to the orifice, the fluid vein will 

w\ v l c j ust fiU I*? an( l tne coefiicient of contraction 
jilB^ will be 
'^fT m t = 1. 

In Fig. 181, if a conicaliy convergent tube 
form the adjutage, the convergent part being of the form 
and length of the vena contracta, and the smallest diame- 
ter of the tube be taken for the orifice, then 

m x = 1. 
392. Coefficients of Velocity.— In Fig. 179, if A x be 
the height to which the jet will rise, then will the velocity 
of discharge be 

v 1 =V2^h 1 ; 
but the theoretical velocity will be 

v = V2gh; 
hence 



[398.] COEFFICIENTS OF DISCHARGE. 295 



\fi* 



h 

from which the value of v x may be determined. 

Or, from the first equation of Article 389, we have 
(writing v x for v), 



v^AG^/, 



2DA' 

by which v x may be computed. 

Let m 2 = — = the coefiicient of velocity, then 

v 

For a mere orifice in a thin plate, . . . m^ = 0.98 
For a short tube, Fig. 182, m 2 = 1.00 

393. Coefficients of Discharge. — The coefiicient of 
discharge is the ratio of the actual discharge to that of the 
theoretical. Let the quantity which flows through an 
orifice, or pipe, or stream be measured, and the quantity 
which should flow be computed ; and let 

Q = the quantity of theoretical flow, 
<2i = the quantity of actual flow, and 



m == the coefiicient of discharge 



then 



»~ > 



But the actual flow equals the mean velocity in the section 
considered into the area of the section ; hence, from Arti- 
cle 380 and the two preceding articles, we have 

Qi = vA 

= m x m 2 .vS 



296 HYDRODYNAMICS. [394) 

which, compared with the preceding value of Q u gives 

From this we find 

m 

ffh = — ; 
mi 

from which the coefficient of the mean velocity may be 
found from the coefficients of discharge and contraction 
For an orifice in a very thin plate, . . m = 0.62 

For a short tube, Fig. 182, m = 0.82 

A comparison of these results shows that the effect of 
the short tube is to reduce the amount of contraction (pro- 
vided there is one in the tube), but that the interference 
of the particles or filaments still reduces the velocity of 
discharge. If a small hole be made in the side of the 
tube, at a distance from the inside of the vessel equal to 
the radius of the orifice, air will rush into the tube, show- 
ing that there is a negative pressure on the tube. If a 
pipe be attached to the tube so as to cover the hole and 
extend down into a vessel of water, the water will rise in 
the tube to balance the negative pressure, the height, 
according to Article 385, being nearly equal to 

(O.lSvf 
2g ; 

in which 0.18 equals 1-0.82. 

394. Large Orifices. — To find the mean velocity of dis- 
charge through a large orifice in the base of a vessel. 

If the orifice is so large compared with the cross section 
of the vessel as to cause a perceptible velocity of the upper 
surface of the liquid, the mean velocity of discharge may 
exceed that due to the head ; for all the particles will have 
an initial velocity which is itself equivalent to a head cor- 



[394.] LARGE ORIFICES. 297 

responding to that velocity. Therefore, the head due to 
the discharge will be the head of the liquid in the vessel, 
jplus the head due to the velocity of the surface. 
Let s be the section of the orifice, 

S ly that of the vessel at the surface of the liquid, 
v and V\, 'he corresponding velocities. 
Then, according to Article 382 we have 

s 

and the head due to this velocity will be, according to 
Article 385, 

,2 



(W 



2? 

which, added to the head h of the liquid, gives, according 
to Article 386, 



v=V^ + k^)> 



from which we readily find 




If s = Si, v = oo ; that is, the velocity must be infiuite 
in order that the section of the issuing vein at the orifice 
shall equal that at the surface of the liquid. 

If s is so small compared with S x that it may be neglected 
then we have 

v = \/2gh y 

as found in Article 383. 
13* 



29S HYDRODYNAMICS. [395.] 

395. External Pressures Considered. 

Let j? be the pressure per unit of area on the issuing 
vein due to the atmosphere or other fluid, and 
j? t , the pressure per unit of area on the upper sur- 
face of the liquid ; 
then the head due to the difference of these pressures will 
be (Article 386), 

which must be added to the head of the liquid ; hence 



*(a + *S*)-|* 



_ P g y h+ f) ~\ 



A discussion of this equation will give several of the 
preceding ones. 

[We cannot follow these modifications further in an elementary 
work, but will add that the formulas have been founded on the 
hypothesis that the velocity of the particles at their exit was gen- 
erated in an infinitesimal of space (Article 383), but it is evident 
that, in a perfect fluid, all the particles in the vessel will be put 
in motion as soon as the liquid begins to flow. If the vessel be 
prismatic, and all the horizontal sections are assumed to remain 
horizontal, and the vessel kept constantly full, the velocity of the 
particles in the upper surface of the liquid being zero, and t the 
time for a particle in the upper surface to reach the orifice ; then 
it is found by means of the Calculus that the velocity of exit 
will be 

TT • Vt 

e -1 , 



Vt 



«2A 

e +1 

in which v has the value given in the preceding equation, *nd e is 

the base of the Naperian logarithms.] 



[396,397.] 



FLOW THROUGH WEIRS 



299 



396. A Weir is an opening in the side of a vessel for 
the discharge of a liquid, in which the upper surface of 
the liquid is &free surface. 

397. A Rectangular Notch. — To find the quantity of 
liquid which will flow from a rectangular notch in the 
side of a vessel, the vessel being kept constantly full. 






-J) 




Fig. 183. 



Fig. 184. 



Let ABOD be the notch, and x the distance of any fila 
raent from the surface BC. The velocity of any fillet 
will be 

v = ^2gx; 
.-. v 2 = 2gx ; 

which is the equation of a parabola, v being an ordinate, 
x an abscissa, and 2g the parameter. Let h = AB, then 
the velocity of the liquid at A will be 

v = <)/2gh. 

In Fig. 184, take AE— V2gh, and construct the para- 
bola BE, then will the velocity at any point in the verti 
cal AB be represented by an ordinate of the parabola 
drawn through that point. The quantity which will flow 
through the orifice in a unit of time will be represented 
by the area of the parabola ABE, multiplied by the width 
ft = BO, Fig. 183 ; hence in a time t it will be represent- 
ed by the area of the parabola multiplied by the breadth 



300 HYDRODYNAMICS. [398-401.] 

of the weir and by the time t. But the area of a parabola 
is two-thirds the area of its circumscribed rectangle ; hence 
we have for the quantity discharged in the time t t 

Q t= m.^VZgJ.h.b.t 

398. The mean velocity of the discharge through a 
rectangular notch at the contracted section will be 

.._ Q 



mbht 



VZgh; 



that is, two-thirds of the maximum velocity through the 
notch. 

399. The coefficient of discharge over a weir depends 
upon the head. If the head is very small the coefficient 
will be small ; but for ordinary cases we have 

m = 0.62 nearly. 

400. Flow through a submerged rectangular ori- 
fice in the side of an upright vessel. 

Let h be the head above the base of the orifice, 
h l7 the head above its upper end, and 
b, its width. 
The discharge will be the same as that due to the differ- 
ence of two weirs having the same breadth b, and heads 
h and h t respectively ; hence the formula of Article 397 
becomes 

Q = imfitV2g(h*—h l *). 

401. Flow in Long Pipes. — The law of resistance in 
long pipes is rather assumed than deduced. It is assumed 
in regard to the velocity, that the resistance due to the 
adhesion of the liquid to the sides of the pipe varies as 



L 401.] 



LONG PIPES 



301 



the square of the velocity, and that due to viscosity varies 
directly as the velocity, so that if a and b are two constants, 
the law will be expressed by 

av 1 + bv. 

In regard to the dimensions of the pipe the resistance will 
vary directly as the length and also as the contour (or 
wetted perimeter) of the pipe, and it is also assumed to 
vary inversely as the area of the cross section of the 
stream. 

If I be the length of the pipe, S the cross section, and c 
the contour, or wetted perimeter, then the law will be ex- 
pressed by 



or 



a e * (tP+fo) 




Fig. 185. 



in which V equals b -f- a. 

The total head will be the head AD = A, of the 
upper reservoir, plus the head due to the fall, or BD sin <£, 
minus the head in the lower reservoir, if there be one. 
The total head, minus the head due to the velocity of dis- 
charge, will equal the total resistance. If the pipe be 
prismatic and full, the velocity will be uniform through- 
out its length. 

Let R = the total head, 

I = DB, the length of the pipe, 
(f> = the inclination of the pipe to the horizontal, 
A == AD, and v = the velocity of discharge ; 
then, if the discharge be into the air, we have 



302 HYDRODYNAMICS. [40aj 

R = h + I sin <f>, 



and 



tf-£=4V+*H* 



Numerous experiments have been made by European 
engineers to determine the constants a and b', among 
which those of Prony, Bossut, and Eytelwein are among 
the most noted. According to the results of these experi- 
ments, D'Aubisson, a French writer, finally wrote the 
equation as follows : 

9 7 ' 

H— |- = 0.000104392 ^ (a 2 + 0.180449a). 

402. Circular Pipes. — The section of pipes being] cir 
cular, if D be the diameter, we have 

8 = ^1?, 
and 

c = irDj 
and the preceding equation becomes 

H— 0.015536a 2 = 0.000417568 i(^+ 0.180449a). 

If the quantity of discharge be given, the velocity may be 
eliminated. 

Let Q = the quantity discharged, then 

/.«= 1.27324 2,; 

which, substituted in the preceding equation, gives 

* For a history of this and other formulae pertaining to the flow of 
water in streams, see Report on the Hydraulics of the Mississippi River 
by Humphreys and Abbott, pp. 207 to 330. 



[403-405.] FLOW IX RIVERS. 303 

H- 0.025187^ = 0.0006769 ~(^+0.141724:§2> 8 ). 

403. To find the diameter of the pipe that will gict 
a given discharge. As 1) is involved to the fifth power, 
it is not practicable to make a direct solution. Omitting 
the terms of least value, that is, the second terms in each 
member, and we have the following approximate value : 

D = 0.2323 J f j£, 

which, for velocities above two feet per second, is consid- 
ered sufficiently accurate. 

404. Condition of the Pipe. — The experiments of 
Mons. Darcy showed that cast-iron pipes, whose interior 
surface was covered with deposits, offered a much greater 
resistance than new and clean cast-iron ones, and that 
when the internal surface was covered with bitumen, or, 
in other words, was practically polished, the resistance was 
least. 

Bends in pipes also diminish the velocity, and sharp 
turns offer much greater resistance than rounded ones. 

405. Flow in Rivers and Canals. — The formula for 
the flow in rivers and canals is of the same general form 
as that given in Article 401 for the flow in long pipes, 
except that when a portion only of the length of the stream 
is considered, and the mean velocity is constant, the head 
due to the terminal velocity will be neglected ; for the 
initial velocity will be the same as the terminal. 

Using the constants which were determined by Prouy 
for this case, we have 

& = 0.00011U15 ^(^+0.217786v): 



304 HYDRODYNAMICS. [406,407.' 

in which c is tne wetted perimeter, that is the line in the 
cross-section which is in contact with the water, and H is 
the fall for the length I. 

Let Q — vS = the quantity of flow, and substituting in 
the preceding equation 

Q 

and omitting the last term, we find 



*& 



Q = 9±.738Sa/- E 



d 



406. Character of the Bed of the Stream. — Mons. 

Darcy found that streams having cement beds offered the 
least resistance to the motion, and that the resistance in- 
creased in the order of the following substances : Cement, 
planks, bricks, gravel, and coarse pebbles. (See Morin'a 
Hydraulique, Troisieme Ed., p. 147.) 

407. Cross Section of the Stream.— It is found by 
observation that the surface of a stream is not horizontal 
in its cross section, but that it is highest where the velocity 




Pig. 186. 



is greatest. This is accounted for by the fact that when 
the fluid is in motion it does not exert as great a side pres- 
sure as when it is at rest ; and as the velocity near the 
shore is very small, it requires a greater head near the 



[408.] 



BACKWATER. 



305 



middle of the stream, where the velocity is greatest, to 
balance the pressure at the sides. It may also be observed 
that, in order to produce a velocity, there will be a greatei 
pressure in the direction of motion in order to overcome 
the resistances to motion, than there will be in a trans- 
verse direction. 

408. Backwater caused by a Dam in a Stream. — 

If a dam be made across a stream, or partly across it, so 
as to elevate the surface at the place of the dam, the sur- 
face of the water above the dam will not be horizontal. 
If a horizontal line OK, Fig. 187, be drawn through the 
crest of the dam, the surface of the water in the pond will 




Pig. 187. 



be entirely above the line, the difference between the 
surface and line being very small at first, but increasing 
gradually as the distance from the dam increases. The 
natural surface may also be elevated for a long distance 
back of the point K, where the horizontal through G in- 
tersects the natural surface of the stream. The elevation 
of the surface above the horizontal CK, including also the 



306 HYDRODYNAMICS. 1409.J 

elevation back of the point K, is called backwater ', and is 
also sometimes called a remou. 

Backwater is caused partly by the inertia of the liquid 
and partly by its viscosity. As the stream approaches the 
dam its velocity is checked, because the pressure on the 
front side of a particle exceeds that on the rear side, and 
when the velocity is thus reduced the particles offer a re- 
sistance to those which succede, and thus the resistance is, 
so to speak, extended up the stream. The resistance due 
to viscosity still further increases this effect. 

Fig. 187 shows a section of the river Weser, in Ger- 
many, at the place of a certain dam, but the horizontal 
scale is much less than the vertical. The mean width of 
the stream was 354 feet, the mean depth about 2.47 feet, 
the depth of the water just above the dam was 9.82 feet, 
and hence the surface was raised 7.35 feet. The slope of 
the stream was quite uniform for a distance of ten miles, 
and averaged 0.000454 per foot. At the point K, where 
the horizontal 67iT, through the crest of the dam, intersected 
the natural surface of the stream, it was found by actual 
measurement that the surface was elevated over 15 inches. 
The distance OK was over three miles. At a distance of 
four miles above the dam, or about one mile above the 
point IT, the elevation of the surface caused by the dam 
was about nine inches. 

In ordinary streams the width, depth, and the character 
of the bed are such variable quantities that the extent of the 
backwater cannot be very accurately computed on a theo- 
retical basis, but empyrical formulas have been given 
which will give an approximate result when applied to 
rivers of about the dimensions of the Weser. (See D'Au- 
bisson's Hydraulics, Article 166.) 

409. Backwater in Rapid Streams. — If the stream is 



[400.1 PROBLEMS. 30? 

very rapid, or rapid compared with its depth, the remous 
will be modified, presenting an appearance similar to that 
shown in Fig. 188. In this case there is a comparatively 
sudden change in the velocity at the head of the pond. 



Fig. 188. 



Problems. 

1 . A prismatic vessel is kept constantly full of a liquid; 
required the time of discharging a given quantity through 
a small orifice in the base. 

Let Q be the quantity, h, the height of the liquid, s, 
the area of the orifice, and t the time ; then 

Q = msvt 
= mstV2gh. 

From this we find the time of discharge to be 

Q 



t = 



ms 



V2gh 



2. Determine the time in which a prismatic vessel will 
empty itself by an orifice in the base. 

Let S be the area of the free surface of the liquid, s 
that of the orifice, x the variable head, v the velocity oi 
discharge, and V the velocity of descent of the free sur- 
face of the liquid. Then 



SOS HYDRODYNAMICS. [409.1 

which, compared with equation (3), page 34, shows that 
the law of descent of the surface is the same as that of 
falling bodies, or rather, it is the same as that of a body 
projected vertically upward ; and from equation (4) on 
page 34 we have 

2A 



in which h is the height to which a body will rise when 
projected vertically upward with a velocity v. Were the 
velocity to remain uniform from the instant that it is pro- 
jected, the time required to go the same distance would be 

.-. t = 2*i ; 

hence the time required for the vessel to empty itself will 
be twice that required to discharge the same quantity when 
the vessel is kept constantly full, If Q be the contents 
of the vessel, we have, by multiplying the last equation of 
the preceding article by two, 

ms\/%gh 

3. To determine the form of a vessel such that the free 
surface of the liquid shall descend uniformly as it dis- 
charges itself through a small orifce at the lower end of 
the vessel. 

Let the vessel be one of revolution, h the height of the 



f409.] PROBLEMS. 309 

upper base above the orifice, r the radius of the upper 
base, x the height of any section above the orifice, y its 

radius, and Ax the thickness of the horizontal laminae ; 
then the conditions of the problem require that the times 
of descent through the small distance Ax shall be the same 
tor all positions of the upper surface. Let s be the area 
of the orifice, and t the time of discharging a quantity 
equal to any lamina, then 

vrisV'Zgh.t = m*. Ax, 
and 

msV^gx.t = Try*. Ax ; 

and dividing one of these equations by the other and re- 
ducing, gives 

which is the equation of a biquadratic parabola. Clep- 
sydras or Water Clocks involve these principles. If the 
time T of the complete discharge of the vessel be given, 
then will 

h _ Ax _ msv2gh _ 
T~~t~ irr 2 ~ C > 

in which c is a constant and is the rate of discharge. From 
this equation we find 

h _ **<* . m 
r 4 ~ %gm l s* ' 

and the equation of the curve becomes 



4 / 2(/?)l 2 8 2 X. 



y - »v 



310 HYDRODYNAMICS. [409.J 

Making x = A, we have, for the radius of the upper base 
of the vessel, 



y = r = y 






EXAMPLES. 

1. A cylindrical vessel, whose height is 3 feet, radius of the 

base 6 inches, is filled with water, and discharges 
itself through an orifice in the base ; if the diameter 
of the orifice is one-half of an inch, and the coefficient 
of discharge, m, is 0.62, in what time will the vessel 
empty itself? 

2. What quantity of water will flow over a weir whose 

breadth is 2 feet and constant depth 10 inches, in 45 
minutes ? 

3. It is required to construct a water clock which will 

empty itself in 10 minutes, the surface descending 
uniformly. The height of the vessel being 24 inches 
and radius of its upper base 3 inches ; required the 
equation of the curve, and the area of the orifice, the 
coefficient of discharge being 0.62. 

4. What quantity of water will, in one hour, flow through 

a pipe 1,500 feet long, 2 inches in diameter, the 
open end being 25 feet below the level of the reser- 
voir? 



EXERGISES. 

1. If a vessel having an orifice in its base be filled successively with 

alcohol, water, and mercury, which will require the least time to 
empty itself, considering each liquid as perfect ? 

2. If a cylindrical vessel be half filled with mercury and the remaining 

half with water, will the velocity of discharge of the mercury 



T409.] EXERCISES. 311 

through an orifice in the base of the vessel be the same as if the 
vessel were entirely filled with mercury ? 

3. If the lower half is water and the upper half mercury, will the flow 

be the same as in the preceding exercise ? 

4. If two liquids of different densities are thoroughly mixed, will the 

velocity of flow from an orifice in a vessel be the same as for each 
liquid separately ? 

5. A block is floating on the water in a vessel, when an opening is sud- 

denly made in the base of the vessel ; considering that the surface 
of the water falls with a decreasing acceleration, will the depth of 
flotation of the block, during the discharge of the water, be the 
same as before the discharge began ? Will the depth of flotation 
remain constant during the discharge ? 

6. If, in a pond which receives no supply, an opening is made in one 

side so that the water will flow out, will the surface remain at a 
true level — that is, parallel to the surface as it stood before the 
opening was made ? 

7. If a vessel filled with water, and having an orifice near its bottom, is 

placed on a platform and made to ascend with a uniform accelera- 
tion, will the velocity of flow through the orifice be the same as if 
the vessel were at rest ? 

8. In Fig. 166, if there be an orifice in the base near the outer edge, 

will the velocity of discharge be the same when the vessel ii 
rotating as when it is at rest ? 

9. When water is flowing uniformly in a pipe is the pressure against the 

aides of the pipe the same as if the discharge be stopped ? 



CHAPTER XXII. 



GASES AND VAPORS. 

410. A Gas is a fluid whose particles are in a constant 

state of repulsion. Common air is taken as a type of gases. 

411. Pressure of the Atmosphere. — If a tube, 32 or 
33 inches long, be closed at one end and filled with mer- 
cury, and the open end be closed with the finger until the 
tube is inverted and the end submerged in a vessel of mer- 
cury, then if the finger be removed the mercury in the 

tube will fall to some point 
B, and remain nearly sta- 
tionary. The column AB 
is sustained by the pressure 
of the atmosphere upon the 
surface of the mercury in 
the vessel A ; and hence the 
weight of a column of mer- 
cury equal to AB, having a 
square inch for its base, will 
equal the pressure of the 
atmosphere upon a square 
inch of surface. The aver- 
age height of the column 
AB at the level of the sea 
is about 29.92 inches (say 30 
inches, or 760 millimeters) 
of mercury, or about 34 feet of water. Hence the mean 
pressure of the atmosphere, at the level of the sea, is about 






Fig. 189. 



[413-414.] MARIOTTE' S LAW. 313 

1 4.7 pounds (say 15 pounds) per square inch. This is called 
the pressure of one atmosphere. The pressure of the at- 
mosphere diminishes as the distance above the earth in- 
creases, and increases for depths below the surface. 

412. The Barometer. — If the tube and vessel shown in 
Pig. 189 be encased in such a way as to retain their rela- 
tive positions while they are carried from place to place, 
and the tube be provided with a suitable scale for reading 
the height of the end B of the mercurial column, above 
the surface A, the instrument is called a barometer. By 
means of it the pressure of the atmosphere may be readily 
determined for any place and at any time. There are 
numerous modifications in the details of different barome- 
ters which, are explained in descriptive works upon the 
subject. 

413. Height of a Homogeneous Atmosphere. — 
If the atmosphere were of uniform density, and the same 
as that at the level of the sea, its height would be found 
by multiplying 30 inches (the height of the mercurial 
column) by the ratio of the density of mercury to that of 
air. Mercury is about 13^- times as dense as water, and 
water 773 times as dense as air when the barometer stands 
at 30 inches ; hence the height would be 

2i x 773 x 13£ = 26088 feet nearly = 5 miles nearly. 
But the actual height is supposed to be from 50 to 100 
miles. This is determined by its effect in deflecting the 
rays of the sun. 

414. Boyle's (or Mariotte's) Law. — Boyle in England 
and Mariotte in France, independently of each other,* 
demonstrated the following law : 

* Writers differ in regard to the respective dates of the discovery 
While some state that both made the discovery at about 1668, other? 
give to Boyle a precedence of several years over Mariotte. 
14 



314 GASES AND VAPOBS. [414.] 

For the same temperature the density of a gas is directly 
proportional to its pressure. 

Both these discoverers proved this law by means of a 
tube, called Mariotte's tube, Fig. 190. Mercury wag 
poured into the tube until the air passage from 
the short to the long tube was just cut off. This 
point was marked zero, and the pressure of the 
air in the short tube was that of one atmosphere 
when the mercury stood at this point. Mercury 
was then poured into the long tube until the air 
in the short column was compressed to one-half 
its length, when it was found that the upper end 
of the long column was about 30 inches above 
the upper end of the mercury in the short tube. 

Again, filling the long tube until the air in the 
short tube is compressed to one-quarter of its 
length, it will be found that the column in the 
long tube above that in the short tube will be 
twice its former length, and so on, observing in 
each case that the temperature must be the same. 
The quantity of air being constant, the density 
will be inversely as the volumes, or directly as the pres- 
sure. 

Let V and V x be the volumes corresponding to pressures 
jp and jpi per square inch, and D and D x their densities; 
then 

L -v± - Q 

Vi P~& 

V 

If V x and p> x are known, the volume V may be found for 




Fig. 190. 



[415,416.] 



MANOMETERS. 



315 



/T\ 



any pressure jp. Considering V and p a6 variables, the 
preceding equation will be of the form 

xy = ra, 

which is the equation of an hyperbola referred to its 
asymptotes. 

415. Boyle's Law is only approximately correct. — 

For many years after the announcement of Boyle's law it 
was confirmed by different experimenters, and the law dur- 
ing that time was supposed to be rigorously correct, but 
more recently the more precise experiments of Despertz 
and Regnault have shown that it differs for different gases 
and is not rigidly true for any gas. But the departure 
from the law is so slight that, for ordinary 
purposes, it may safely be considered as 
exact. 

416. Manometers are instruments for 
measuring the tension of a gas. The ten- 
sion is the pressure per square inch, and is 
often compared with the pressure of one 
atmosphere. The principle of the mano- 
meter is founded on Boyle's law of the com- 
pressibility of gases. There are many kinds, 
but the closed manometer, Fig. 191, is one 
of the most common. It consists of a ver- 
tical tube closed at the upper end, the lower 
end opening into a vessel of mercury or 
other liquid. Another tube connects the 
liquid with the vessel containing the gas, so 
that the pressure of the gas when it is ad- 
mitted into this tube will force the liquid up 
the long one, thereby compressing the air above the liquid. 
The vertical tube is provided with a scale for indicating 




Fig. 191. 



316 GASES AND VAPORS. [417-419.] 

the pounds of pressure per square inch, or the number of 
atmospheres, or both, as may be desired. 

417. Expansion of Gases due to a Change of Tem- 
perature. — Reliable experiments show that the expansion 
of all gases under constant pressure may be considered as 
the same for each degree of increase of temperature for 
all ranges of temperature. Still the delicate experiments 
made by Regnault show that the expansions are not identi- 
cally the same, and that the increase of volume increases 
somewhat more rapidly than the increase of temperature. 

Assuming that Mariotte's law is rigorously exact, and 
that the rate of expansion of a gas is the same as that of 
the increase of its temperature, it follows that the tension 
of a gas under constant volume varies directly as the 
change of temperature. 

418. Coefficient of Expansion due to Temperature. 

— The volume of dry air under a constant pressure in- 
creases 0.002039 of its volume at 32° and 14.7 lbs. for 
each increase of 1° Fahrenheit, and this is called the 
coefficient of expansion. It is also considered as the 
coefficient of tension under a constant volume for each 
1° Fahrenheit. 
Let a be the coefficient of expansion (or tension), then 

a = 0.002039 for 1° F. 
= 0.003670 for 1° G. 

419. To find the Volume of a Gas due to a change 
of Temperature and Pressure. 

Let V ,po, t , be the initial volume, tension, and temper 
ature, and V u p u t u the corresponding terminal values 
Then will the change of temperature be 

i\ — Iq> 
and hence the volume due to this change will be 



F420.J EXPANSION OF GASES. 317 



r 1= :(l +<•&-*)) P.; 



and, according to Mariotte's law, if there be a change of 
pressure at the same time the volume will be 



Fi = (l+«( 4 _4))JJF iJ 

H i+ ^>)§- 






Let V 2 ,j> 2 , t%, be the quantities for another pressure and 
temperature; then 

Dividing the former equation by the latter, gives 

V i = l+a(t i -t ) ^p^ 

V 2 l-ha(t 2 —t ) jpx 
420. Perfect Gas. — The more rare a gas is the more 
perfect it is considered to be. A perfect gas is defined to 
be one which is destitute of mass. A perfect gas does not 
exist, but this ideal gas serves as a standard of comparison 
for different gases. It ma) 7 be defined to be the limit 
towards which gases approach as they are expanded indefi- 
nitely. The limit towards which the coefficient of expan- 
sion approaches is not definitely known, but it is assumed 
by Kankine to be 

a = 0.0020275 = -J— for 1° F. 

= 0.00365 = ^- for 1° O. 

274 

and these are called the coefficients of expansion for a 
perfect gas. 



318 GASES AND VAPORS. [421.) 

Substituting the former of these in the formula above, 
and taking the initial temperature at the melting point of 
ice, in which case t = 32° F., we have 

493.2^= (461.2 +A)p^. 

If t\ were taken at 461.2° F. below zero, this expression 
would vanish. The ideal temperature —461.2° F., or 
— 493.2° F. below melting ice, is called the absolute zero. 
This temperature cannot be even approximately reached 
by any known process ; but, according to the formula, it 
is a temperature at which a perfect gas would lose all ex- 
pansive power. 

The absolute zero is used because the formulas for the 
expansion of gases due to temperature are simplified when 
the temperature is reckoned from that point 

421. Thermometers are instruments for measuring 
changes of temperature. They are made on the principle 
of the uniform rate of expansion of liquids. Mercury is 
most commonly used, but alcohol is sometimes used ; and 
the latter is necessarily used instead of mercury for tem- 
peratures below —40° F., for mercury freezes at about 
that temperature. Three scales are used, viz. : Fahren- 
heit's, marked F., the Centigrade, marked C, and Reau- 
mur's, marked R. 

Fahrenheit chose for the zero of his scale the height of 
the mercury, which was produced by a mixture of salt and 
ice. This he believed to be the absolute zero of cold. The 
height produced by the boiling point of water he marked 
212, and divided the space between these points intc equal 
parts. The freezing point of water was 32 divisions above 
^ero. 

The zero of the Centigrade thermometer is at the frees- 



1422,423.] THERMOMETRIC SCALES. 319 

ing point of water, and the boiling point of water is marked 
100, and the space between is divided into equal parts. 

In Reaumur's scale the zero is fixed at the freezing 
point of water, and the boiling point of water is marked 
SO, the space between being divided into 80 equal parts. 

The melting point of ice is now used instead of the 
freezing point of water, for the latter is not constant. In 
all thermometers the divisions below zero are considered 
minus. 

422. To Convert one Thermometric Scale into 
another. — The number of divisions between the melting 
point of ice and the boiling point of water on the three 
scales is 

F. C. R. 

180, 100, 80 : 



or as 



1, 



5 4 

9' 9 



But the Fahrenheit scale begins 32° below the others ; 
hence, if F°, C°, and R° represent the degrees on the 
respective scales for the same temperature, we have 

C°=: | (F°-32°) ; R°= \ (F°-32°) ; 

from which we find 

F°=f C°+32°, F°=jR° + 32°, andC =5R°. 
5 4 4 

423. Compressed Air. — Air, compressed to several 
atmospheres, may be used instead of steam for driving 
engines. If the engine is at a great distance from the 
compressing machine it is a more desirable power than 



320 GASES AND VAPOES. [423.] 

steam, for steam will condense and thus lose its power, 
but compressed air may be conducted for miles, if neces- 
sary, without any loss of power except that due to the 
leakage and friction of the pipes in which it is con- 
ducted. It is especially serviceable for driving engines 
underground. It is the chief power for driving engines 
in the construction of large tunnels and in underground 
mining;. It was thus used in the construction of the 
Mont Cenis tunnel in Europe, and the Hoosac tunnel 
in this country. 

When air is suddenly compressed heat is developed, and 
sometimes the heat becomes so intense as to make the 
compressor very hot. If this heat is lost while the air 
is passing from the compressor to the engine, power is 
lost. Methods, therefore, have been devised for keeping 
the air as cool as possible during compression. The most 
practical way is to inject a spray of water into the com- 
pressor while the air is being compressed. If the heat 
that is in the air when it leaves the compressor could be 
maintained without extra cost, until the air is used, there 
would be no loss due to the development of the heat, and 
as the air thus heated has a greater tension, it would be 
undesirable to reduce the temperature during compression 
f>o far as this cause is concerned. 

[The following formula gives the relation between the pres- 
sure, density, and temperature of air during rapid changes of 

motion. (See The Author's Analytical Mechanics, p. 381) : 



H*f=(£)' 

8.-W' 



in which ti is the absolute temperature of the air (or gas) when 
the pressure is pi and density §i, and t 2 , p<2, 8 2 , the corresponding 
quantities in another condition of the gas. y is the ratio of spe» 



f*24.] 



STEAM OR YAPOE. 



321 



cific heat of constant pressure to the specific heat of constant 
volume, which for a perfect gas is 1.408, which value is suffi- 
ciently exact for dry air. The quantity of air being constant, the 
volumes will be inversely as the densities ; hence we have 



Tj \v 2 J 



Let the volume Vi, at 14.7 pounds, be 100, and the tempera- 
ture 60° F. , then will the absolute temperature be t x = 521.2°, 
If the air be suddenly compressed, we will have, according to 
this formula : 



Pressure, lbs. per 
square inch. 


Resulting temper- 
atures. 
Deg. F. 


Resulting 
Volumes. 
(Relative.) 


Relative Volumes 
under constant 
temperature. 
(Boyle's Law.) 


14.7 


60 


100 


100 


30.0 


180 


60 


49 


45.0 


259 


45 


32 


60.0 


322 


37 


25 


75.0 


374 


31 


20 


90.0 


420 


28 


16 


105.0 


460 


25 


14 



After passing the compressor, if the heat escapes so as to re- 
duce the temperature to 60°, the volumes will be reduced from 
those in the third column to those in the fourth, and there will 
be a corresponding loss of tension in the air. (For results of 
computations and experiments to determine the work lost in 
using compressed air, see Kent's "Mechanical Engineer's 
Pocket-book." 

424. Steam or other vapor, separated from liquids, may 
follow the same laws in regard to expansion, temperature, 
and density, as air and other gases. If, however, the steam 
be in contact with the water from which it is generated, 
the temperature cannot be increased without, at the same 
time, increasing both the tension and density of the steam, 
14* 



322 GASES AND VAPORS. [421j 

Steam, in this condition, is, for a given temperature, al- 
ways at its maximum tension, and also at its maximum 
density. It may be said to be constantly at its dew point. 
Steam in which both the density and tension change on 
account of a change of temperature is called saturated 
steam. But steam in which the tension may change with 
the temperature without changing its density is called 
steam gas. . It follows the laws of permanent gases in this 
respect. The tension of such steam may greatly exceed 
that of saturated steam for high temperatures, and when 
thus heated it is called Superheated Steam. 

Problems. 

1. To find the weight of a cubic foot of air at any temr 
perature and pressure. 

The weight of a cubic foot of air at 32° F., when the 
barometer is at 30 inches, is 0.08072 pounds avoirdupois. 
For a given mass of air the weight of a given volume will 
be inversely as the temperature and directly as the pres- 
sure, or generally, inversely as the total volumes of the air 
under the different conditions. Hence, the weight, W l9 of 
a cubic foot will become, according to Article 419, 
W 1 _ Vp _ 1 p, 

W V t 1+0.002039(^-32°) '30 

ii7 _ 0.08072^ 



28.0425+ 0.061170* ' 
in which p x is the reading of the barometer. 

2. Find the weight of a cubic foot of steam at any tern* 
perature and pressure. 

The density of steam is five-eighths of the density of air 
for the same tension and temperature; hence, when the 
pressure is given in pounds per square inch, we have 



[424.J 



PEOBLEMS. 



323 



W = 



f x 0.08072 



Pi 



1 + 0.002039(^-32°) '14.75 
0.05045^ 



13.7877+ 0.030075*" 

3. A spherical air-bubble rises vertically from a depth 
h to the surface of the liquid ; required the equation of 
the curve described by the extremities of a horizontal 
diameter. 

Let r = the radius at the surface, 
y = the radius at a depth x. 

The volumes will be inversely as the pressures. At the 
surface the pressure per unit is represented by a column 
of water 34 feet high, and at a depth x by 34+ x; hence, 

frr/* : frn/ 8 : : 34 + ^:34; 

34r* 



'.*» = 



34+a?' 



or 



— 34(£-l). 



r~\ 



4. Required the degree of exhaustion from the receiver 
of an air-pump. 

At each stroke of the piston a 
quantity of air is removed from the 
receiver B and the pipe a c equal to 
the volume in either barrel B or B', 

Let V be the volume in the re- 
ceiver and pipes, v, the contents of 
either barrel, D , the initial density, 

A» D% D n , the densities after 

the successive strokes. 



zsr 



Fig. 192. 



^ r - 



324 GASES AND VAPORS. T4£l; 

The quantity remaining after the first stroke will be 

J? V-B Q v, 

which will be uniformly distributed throughout the re« 
ceiver and pipes, and the density will be reduced to D x ; 
hence 

D 1 V=D V-Z> v; 

/.A=A(i~). 

Similarly, at the end of the second stroke, we have 

A ■ = A(l-f) = A(l- £ )'; 

and at the end of the n th stroke 

The density D n cannot be zero so long as n is finite ; 
hence the exhaustion can never be perfect. Indeed, the 
degree of exhaustion falls far short of that indicated by the 
formula, for the valves and pistons cannot be made per- 
fectly air-tight, and it requires some pressure to operate 
the valves, so that after a few strokes the exhaustion, in 
practice, proceeds very slowly. 

5. To determine elevations by means of a barometer. 

In order to solve this problem it is necessary to know 
the law of diminution of the pressure of the atmosphere. 
Consider three consecutive strata of the atmosphere of 
equal thicknesses, but so thin that the density of each 
may be considered uniform. The pressure on the top of 
the upper stratum will be the weight of the atmosphere 
above it, which, if it is near the level of the sea, will be 
nearly 14.75 pounds. Let this pressure be j> . The pres- 



[424.] PROBLEMS. 325 

sure upon the second stratum from the top will bejpb, 
plus the weight of the stratum ; hence, 

p l —p = the weight of the first stratum. 

Similarly, 

P2 —P\ = the weight of the second stratum. 

The weights of the strata are as their densities, or 

JPi—JPo : 1?2-Pi : : A : A- 
But, according to Mariotte's law, the densities vary directly 
as the pressures ; hence, 

These quantities follow the law of a geometrical pro- 
gression ;* hence, the natural numbers, 1, 2, 3, etc., which 
are as the numbers of the successive laminae, will be the 
logarithms of the successive pressures. But the system of 
logarithms remains to be determined. The thicknesses of 
the laminae being constant, the distance below the initial 
point will equal the thickness of a lamina into the number 
of laminae. 

* Let a be the first term of a geometrical series, 
r, the ratio, and 
n, the n th term ; 
then for three consecutive terms we have 

then if ar» be substituted for p , ar n +i for p lf and fl^+Sforpa, the 
expression in the text will reduce to the identical equation 

r ~~ r' 



326 GASES AND VAPORS. [434., 

Let Aa = the thickness of a lamina, 
2 = the number of laminae, 
h = z.Aa = the distance from the initial point, 
j> = the pressure at the initial point, 
j) = the pressure at the distance h from the initial 

point, 
a = the base of the system of logarithms, and 
m and n constants to be determined ; 

then the form of the expression for the law of pressurei 
will be 



lo g«^ =7 ^; 



which may be written 



m 



<p = ma 1 *. 



Thus far the distance has been reckoned downward. If 
it be reckoned above the initial point, the sign of h will bo 
changed, and we have 

p = mar 1 *. 

In this equation, if h = 0, the pressure will be represented 
by p ; hence, 

and the equation becomes 

p —^ar 1 *. 

Let b be the reading of the barometer at the initial 
station and b the reading at the second station, then 

b _jp 

and the preceding equation becomes 

b = boa- 1 * ; 
hence, taking the logarithms of both sides, we find 



T434 PEOBLEMS. 397 

in which the subscript, a, indicates that the base of this 
system is. a. To find n and & requires at least two obser- 
vations at cwo known elevations ; * but, without attempting 
to find them in this place, we observe that it has been 
found that the base is that of the Naperian system of 
logarithms, and 

n 

m These relations are easily deduced in the following manner : Let p 
bft the pressure due to one atmosphere, whose height is H and densitf 
D, an£ dp the pressure due to a lamina whose thickness is dx. Then 

p = Dff i 
and 

dp = Ddx. 

Dividing the latter by the former gives 

dp _dx 

integrating which between the limits of p and p, and and a?, give* 

logp -logi>=J:; 

or 

t P x . 

,\p=ptf b=Po6 f'i 



p 



3^8 GASES AND VAPORS. [424.] 

the height of a homogeneous atmosphere, which, according 
to Article 413, is 

E= 26,170 feet nearly. 

If common logarithms are used, they must be divided by 
the modulus of the common system ; or, what is the same 
thing, multiplied by the reciprocal of the modulus, to re- 
duce the result to an equivalent value. Hence, the value 
of h becomes 

h =2.30258x26170 log ^ 

= 60258 log ^. 

By means of actual observations it has been found that the 
coefficient should be somewhat larger than that given 
above, and that 

h = 60346 log -|° 

gives better results. 

It is necessary, however, to add several corrections to 
this formula. The pressure of the air and the weight of 
the mercury will both vary with the force of gravity. The 
force of gravity at any latitude, Z, that at 45 degreel 
being unity, will be (see page 32), 

. 0.08238 T 
* = 1 - 321726 C0S2Z 

= 1-0.00256 cos 2 Z, 

and the coefficient 60346 must be multiplied by this 
quantity. 

The density of the atmosphere also changes with its 
temperature. Let t x be the temperature at the iirst sta- 



T424.] PROBLEMS. 329 

tion and t 2 that at the second, then will the mean temper- 
ature be 

ifa + t*), 
which may be considered as the temperature above or be- 
low 32°. Hence, the expansion of the air will be repre- 
sented by the expression 

l + a[^ 1 + 4)-32°] 

= l + 0.002039[i(^ + 4)-32 o ] ; 
or, for the Centigrade scale, 

l + 0.00366[^ 1 + £ 2 )]. 

Hence, the formula for the height becomes 
A (in feet) = 60346(1-0.00256 cos 2£)[l+0.00102(* 1 +* 2 -64°)]log^-\ 

which is the formula given by Laplace. But Poisson, in 
his Traite de Mecanique, pages 622-636, introduces cor- 
rections for the expansion or contraction of the mercurial 
column due to changes of temperature, as determined by 
an attached thermometer. (Mercury expands 0.0001001 
of its length for each degree F. of increase of its tempera- 
ture). He also gives corrections for the diminution of 
gravity due to an elevation. (It varies inversely as the 
square of the distance measured from the centre of the 
earth.) Also a correction due to the attraction of a moun- 
tain when observations are taken near it. It is rare, how 
ever, that all these refinements are used in practice. 

At the level of the sea the mercury stands at 30 inches nearly. 

6,000 feet above " " 24.7 " 

10,000 feet (height of Mfc. ./Etna) " 20.5 " 

t5,000 feet (height of Mt. Blanc) " 16.9 " 

3 miles ....... 16.4 " 

6 miles 8.9 '• 



330 GASES AND VAPORS^. ^424.] 

6. To find the velocity of discharge of ah into a 
vacuum through a small orifice in a vessel, the pressure 
within the vessel being equal to one atmosphere. 

This may be solved according to three different hypo- 
theses : 

1st. Suppose that the gas is incompressible. In thia 
case the head due to the velocity will equal the height of 
a homogeneous atmosphere, and we have 

v =V2g~h =V64:ix 26170 
= 1300 ft. per sec. nearly, 
= 886 miles per hour nearly. 

2d. Suppose that the gas is compressible and perfectly 
elastic. Then it may be shown by higher analysis that 

* = V2i>log|-, 

in which P denotes the pressure of one atmosphere^ the 
pressure within the vessel, and p t the external pressure. 
But when the issue is into a vacuum^ = o, and we have 

v = 00, 

that is, for this case, according to this. hypothesis, the velo- 
city will be infinite, a result which is incorrect, as shown 
in the following Article. 

3d. Consider the gas not only as elastic and compress^ 
ble, but also that its temperature and density change 9ud 
denly as the gas escapes. It may be found for this rose 
(see Eankine's Applied Mechanics, page 582) that 

*-(F^L w y 

ro which 



[434.: PBOBLEMS. 331 

7 = 1.408, 

t = 493.2° F. above absolute zero = the absolute 
temperature of melting ice, 

t == the absolute temperature of the gas in the 
vessel, 

Pq = the height of the mercurial column at the level 
of the sea, 

Sq = the density of air compared with that of mer- 
cury, 

p 2 = the pressure per unit against the issuing jet, 

and 
j) t = the pressure per unit within the vessel. 
If the jet issue into a vacuum, we have 

and the equation becomes 



L(7-l)VoJ' 



■£? = 26,214, as given by Rankine, which is the height of 
a homogeneous atmosphere for dry air. 
Substituting the several quantities given above in the 

preceding equation gives 



\\T± 



v = 2,413y — feet per second. 

TO 

At the temperature of freezing, r x = t , and we have 
v = 2,413 feet per second. 

EXAMPLES. 

1. Required the number of degrees through which a given 
volume of air must be heated so as to double its 
volume, the pressure remaining constant. 

Am, 490. 



332 GASES AND VAPORS, [424.] 

2. A tube 30 inches long, closed at one end and open at 
the other, was sunk in the sea with the open end 
downward, until the inclosed air occupied only one 
inch of the tube ; what was its depth ? 

Ans. 986 feet. 

A spherical air-bubble having risen from a depth cf 
1,000 feet in water, was one inch in diameter when 
it reached the surface ; what was its diameter at the 
point where it started ? Ans. 0.32 inches. 

A balloon whose capacity is 10,000 cubic feet is filled 
with hydrogen gas ; the specific gravity of the gaa 
being 0.0690 compared with air, how many pounds 
of ballast will just prevent it from rising ? 

Five cubic feet of air at 32° F. and pressure of 15 lbs. 
per square inch, is confined in a vessel ; what will be 
the tension when heated to 400° F. and the volume 
increased to 5.5 cubic feet ? 



The following examples are selected from the London 

University Examination Papers, from 1862 to 1870 : 

1. Explain the difficulty of opening a lock-gate when the water is at a 

different level within and without the lock ; also, why the force 
required to open the gate is not proportional directly to the differ- 
ence of level. 

2. The weight of water is 770 times that of air ; at what depth in a 

lake would a bubble of air be compressed to the density of water, 

supposing the law of Mariotte to hold good throughout for the 

compression ? 
8. A body weighs in air 1,000 grains, in water 300 grains, and in another 

liquid 420 grains ; what is the specific gravity of the latter liquid ? 

Ans. .8285. 
4. A mercurial barometer is lowered into a vessel of water, so that the 

surface of the water is finally six inches above the cistern of the 



[424.] EXAMPLES. 333 

barometer. What kind of change will take place in the reading 
of the column of the instrument ? Give a reason for your reply. 

5. If a bottle filled with air be tightly corked, and lowered into the 
ocean, the cork will be forced in at a certain depth. Why is this ? 
and what will take place if the bottle be filled with water instead 
of air? 

6 If a barometer were carried down in a diving-bell, what would take 
place ? 

7. A solid, soluble in water but not in alcohol, weighs 346 grains in air, 

and 210 in alcohol; find the specific gravity of the solid, that of 
alcohol being 0.85. Ans. 2.1625. 

8. A body whose specific gravity is 3.5, weighs 4 lbs. in water. What 

is its real weight ? Ans. 5.6 lbs. 

9. If as much additional air were forced into a closed vessel as it pre- 

viously contained when in communication with the atmosphere, 
what would be the pressure on a square inch of the internal 
surface ? 

10. At what depth in . a lake is the pressure of the water, including the 

atmospheric pressure, three times as great as at the depth of 10 
feet, on a day when the height of the liquid column in a water- 
barometer is 33 feet 6 inches ? Ans. 97 feet. 

11. A lump of beeswax, weighing 2,895 grains, is stuck on to a crystal 

of quartz weighing 795 grains, and the whole, when suspended in 
water, is found to weigh 390 grains ; find the specific gravity of 
beeswax, that of quartz being 2. 65. Ans. . 965. 

12. A barometric tube of half an inch internal diameter is filled in the 

usual way, and the mercury is found to stand at the height of 30 
inches. A cubic inch of air having been allowed to pass into the 
vacuum above the mercury, the column is found to be depressed 
5 inches. What was the volume of the original vacuum ? 

13. A bottle holds 1,500 grains of water, and when filled with alcohol it 

weighs 1,708 grains ; but when empty it weighs 520 grains ; what 
is the specific gravity of alcohol ? Ans. .792. 

14. A piece of cupric sulphate weighs 3 ozs. in vacuo, and 1.86 ozs. in 

oil of turpentine ; what is the specific gravity of cupric sulphate, 
that of turpentine being 0.88 ? Ans. 2^§. 

16. If the height of the barometer rises from 30 inches to 30.25 inches, 
what is the increase of pressure (in ozs.) upon a square foot ? — the 



334 GASES AND VAPORS. [424.] 

weight of a cubic foot of water being taken to be 1,000 ozs., and 
the specific gravity of mercury 13.56. Ans. 282.5 ozs. 

It A cylindrical vessel standing on a table contains water, and a piece 
of lead of given size supported by a string is dipped into the 
water ; how will the pressure on the base be affected (1) when the 
vessel is full, (2) when it is not full ? and in the second case, 
what is the amount of the change ? 

17. A wooden sphere has a small hole drilled in it, and is placed in 

water. Find its positions of equilibrium ; and state which position 
is of stable, and which is of unstable equilibrium. 

18. The water above the empty lock of a canal is 8 feet higher than the 

base of the floodgates, which are 4 feet broad, and provided with 
handles 10 feet long ; find what force would have to be applied 
to the extremity of the handle to force open a floodgate, without 
previously letting in the water, assuming a cubic foot of water to 
weigh 1,000 ozs. avoirdupois. Ans. 1,600 lbs. 

19. A balance is wholly immersed in water, and a body appears to 

weigh 1 lb. , the weights against which it is balanced having the 
specific gravity 8.5. What will it appear to weigh when balanced 
against weights of the specific gravity 11.5 ? -j^g 

Ans.— lbs. 

20. When a body is floating partly immersed in a liquid, what effect 

will a fall of the barometer have upon the body ? 

21. The specific gravity of cast copper is 8.79, and that of copper wire 

is 8.88, What change of volume does a kilogramme of cast copper 
undergo in being drawn out into wire ? 

Ans: 1.15 cubic centimetres. 

22. A cylindrical wooden rod of specific gravity .72 and 1 centimetre in 

diameter is loaded at one end with 9.08 grammes of lead (specific 
gravity = 11.35) ; how long must the rod be in order that it may 
just float in water at the maximum density ? 

23. State the relation between the pressure and density of an elastic 

fluid. 

24. A piece of cork floats in a basin of water, and the basin is placed 

under the receiver of an air-pump. State and explain the effect 
of pumping out a portion Of the air in the receiver. 

25. A wineglass is inverted, and its rim just immersed in water. What 

would be the effect of placing a small piece of ice in the water be 
neath thejglass ? 



[424.] EXERCISES. 33o 

26. Find the atmospheric pressure on a square inch, assuming that the 

height of a column of water supported by the atmospheric pres- 
sure is 30 feet, and that a cubic fathom of water weighs six tons 

27. Compare the depths to which a right cone must be immersed in a 

fluid of twice its density, that it may be in equilibrium when (1) 
the vertex is downwards, and (2) the base. 
38. A floodgate is 6 feet wide and 12 feet deep. Reckoning the weight 
of a cubic fathom of water at 6 tons, what is the total pressure on 
the floodgate when the water is level with its top ; and what ia 
the situation of the centre of pressure ? 

29. A cubic inch of one of two liquids weighs a grains, and of the other 

b grains. A body immersed in the first fluid weighs p grains, and 
immersed in the second fluid weighs q grains. What is its true 
weight, and what is its volume ? 

Ans. V = (p-q)-^(b-a), W = {bp—aq)-t-(b—a). 

30. A quantity of air contained in a spherical vessel is transferred first 

into a cylindrical vessel, and then into a cubical vessel, each of 
which would just circumscribe the spherical vessel. Compare the 
total pressure produced by the air on the walls of the three vessels, 

Ans. 5*» 2 : 12* 2 : 192. 



APPENDIX, 



TABLE I. 

EXPERIMENTS ON FRICTION, WITHOUT UNGUENTS. BY M. MORTN. 





Friction of 
Motion. 


Feiction of 
Quiescence. 


SURFACES OF CONTACT. 


II 

is « 


.s°§ 

.S jd -3 

sirs 


8 2 

■ O aj 

O o 


Sal 


Oak upon oak, the direction of the fibres 


0-478 

0'324 
0-336 

0-192 


25° 33' 

17 58 

18 35 

10 52 



0-625 
0-540 

0-271 
0*43 


82° 1' 


Oak upon oak, the direction of the fibres 
of the moving surface being perpen- 
dicular to those of the quiescent sur- 
face and to the direction of the mo- 
tion* 


28 23 


Oak upon oak, the fibres of both surfaces 
being perpendicular to the direction of 
the motion 




Oak upon oak, the fibres of the moving 
surface being perpendicular to the 
surface of contact, and those of the 
surface at rest parallel to the.direction 


15 i0 


Oak upon oak, the fibres of both surfaces 
being perpendicular to the surface of 
contact, or the pieces end to end 


23 17 



* The dimensions of the surfaces of contact were in this experiment *947 square feet, 
and the results were nearly uniform. When the dimensions were diminished to -043, a 
tearing of the fibre became apparent in the case of motion, and there were symptoms of 
the combustion of the wood ; from these circumstances there resulted an irregularity in 
tne friction indicative of excessive pressure. 

15 



338 ELEMENTARY MECHANICS- 

TABLE I.— Continued. 



SURFACES OP CONTACT. 



Elm upon oak, the direction of the fibres 
being parallel to the motion 

Oak upon elm, ditto * 

Elm upon oak, the fibres of the moving 
surface (the elm) being perpendicular 
to those of the quiescent surface (the 
oak) and to the direction of the mo- 
tion 

Ash upon oak, the fibres of both surfaces 
being parallel to the direction of the 
motion 

Fir upon oak, the fibres of both surfaces 
being parallel to the direction of the 
motion 

Beach upon oak, ditto 

Wild pear tree upon oak, ditto 

Service tree upon oak, ditto 

Wrought iron upon oak, ditto + 

Wrought iron upon oak, the surfaces be- 
ing greased and well wetted 

Wrought iron upon elm 

Wrought iron upon cast iron, the fibres 
of the iron being parallel to the motion 

Wrought iron upon wrought iron, the 
fibres of both surfaces being parallel 
to the motion 

Cast iron upon oak, ditto 

Ditto, the surfaces being greased and 
wetted 

Cast iron upon elm 

Cast iron upon cast iron 

Ditto, water being interposed between 
the surfaces 

Cast iron upon brass 



Fbiotion op 
Motion. 






0-432 
0-246 



0-450 



0-400 



0-355 
0-360 
0*370 
0-400 
0-619 

0.256 
0-252 

0-194 



0-138 
0-490 



0-195 
0-152 



0-314 
0-147 



§■21* 



23° 22' 
13 50 



24 16 



21 49 



19 33 

19 48 

20 19 

21 49 
31 47 

14 22 

14 9 

10 59 



7 52 

26 7 



11 3 

8 39 



17 26 
8 22 



Fbiotion of 
Quiescence. 






0-694 
0-376 



0-570 

0-570 

0-520 

0-53 

0-440 

0-570 

0-619 

0-649 

0-194 

0-137 

0-646 
6 : 162* 



111 



34° 46* 

20 37 



29 41 



29 41 



27 29 

27 56 

23 45 

29 41 

31 47 



10 59 

7 49 

32 52 
*9**i3 



* It is worthy of remark that the friction ' oak upon elm is but five-ninths of that of 
elm upon oak . 

t In the experiments in which one of the si rfaces was of metal, small particles of tha 
metal began, after a time, to be apparent upon the wood, giving it a polished metallic 
appearance ; these were at every experiment wiped off ; they indicated a wearing of tha 
metal. The friction of motion and that of quiescence, in these experiments, coincided. 
The results were remarkably uniform. 



APPENDIX. 
TABLE I. — Continued, 



8UBPACES OF CONTACT. 



Oak upon cast iron, the fibres of the 
wood being perpendicular to the direc- 
tion of the motion 

Hornbeam upon cast iron — fibres parallel 
to motion 

Wild pear tree upon cast iron — fibres 
parallel to the motion 

Steel upon cast iron 

Steel upon brass 

Yellow copper upon cast iron 

Do. do. oak 

Brass upon cast iron 

Brass upon wrought iron, the fibres of 
the iron being parallel to the motion. . 

Wrought iron upon brass 

Brass upon brass 

Black leather (curried) upon oak * 

Ox hide (such as that used for soles and 
for the stuffing of pistons) upon oak, 
rough 

Do. do. smooth 

Leather as above, polished and hardened 
by hammering 

Hempen girth, or pulley-band (sangle de 
chanvre), upon oak, the fibres of the 
wood and the direction of the cord be- 
ing parallel to the motion 

Hempen matting, woven with small 
cords, ditto 

Old cordage, 1% inch in diameter, ditto 

Calcareous oolitic stone, used in build- 
ing, of a moderately hard quality, 
called stone of Jaumont — upon the 
same stone 

Hard calcareous stone of Brouck, of a 
light gray color, susceptible of taking 
a tine polish (the muschelkalk), mov- 
ing upon the same stone 



Fbiction of 
Motion. 



Friction op 
Quiescence. 



43 a 

a 2 

O o 



0-372 

0-394 

0-436 
0-202 
0-152 
0-189 
0-617 
0-217 

0-161 
0-172 
0-201 
0-265 



0-52 
0-335 

0-296 



0-52 

0-32 
0-52 



0-64 



0-38 



Zi Jii at 

gel 



20° 25' 

21 31 

23 34 

11 26 
8 39 

10 49 

31 41 

12 15 



9 46 
11 22 
14 51 



27 29 

18 31 

16 30 



27 29 

17 45 

27 29 



32 38 



20 49 



43 fl 

c o 
'S.2 

O O 



0-617 



0-74 



0-605 
0-43 



0-64 



0-50 
0-79 



0-74 



0-70 



■ni 



81 41 



36 31 



31 11 
23 17 



32 38 

26 34 
38 19 



36 31 



35 € 



* The friction of motion was very nearly the same whether the surface of contact wal 
luside or the outside of the skin. The constancy of the coefficient of the friction of 
motion was equally apparent in the rough and the smooth skins. 



&40 ELEMENTARY MECHANICS. 

TABLE I.— Continued. 





Friction op 
Motion. 


Friction op 

Quiescence. 


SURFACES OF CONTACT. 


0.0 
Br 1 
o w 
O o 


b0«H | 

nog 


■p a - - 
a o 

'3.2 

O u_i 

O o 




The soft stone mentioned above, upon 


0-65 
0-67 
0-65 

0-38 
0-69 
0-60 
0-38 
0-84 


33° 2' 
33 50 

33 2 

20 49 

34 37 
30 58 
20 49 
13 30 


0-75 
0^75 
0-65 

0-63 
0-49 
0-67 
064 
0-43 


36° 53' 


The hard stone mentioned above upon 
the soft 


36 53 


Common brick upon the stone of Jau- 
mont 


33 2 


Oak upon ditto, the fibres of the wood 
being perpendicular to the surface of 
the stone 


32 13 


Wrought iron upon ditto 

Common brick upon the stone of Brouck 

Oaksas before (endwise) upon ditto 

Iron, ditto ditto .... 


26 7 
33 50 
32 38 
22 4? 



APPENDIX. 



341 



TABLE H. 

EXPERIMENTS ON THE FEICTION OF UNCTUOUS SURFACES. 
BY M. MORIN. 

In these experiments the surfaces, after having been smeared with an 
nnguent, were wiped, so that no interposing layer of the unguent prevented 
their intimate contact. 



SURFACES OF CONTACT. 



Oak upon oak, the fibres being parallel 
to the motion 

Ditto, the fibres of the moving body be- 
ing perpendicular to the motion 

Oak upon elm, fibres parallel 

Elm upon oak, ditto 

Beech upon oak, ditto 

Elm upon elm, ditto 

Wrought iron upon elm, ditto 

Ditto upon wrought iron, ditto 

Ditto upon cast iron, ditto 

Cast iron upon wrought iron, ditto 

Wrought iron upon brass 

Brass upon wrought iron 

Cast iron upon oak, ditto .... 

Ditto upon elm, ditto, the unguent being 
tallow 

Ditto, ditto, the unguent being hog's 
lard and black lead 

Elm upon cast iron, fibres parallel 

Cast iron upon cast iron 

Ditto upon brass 

Brass upon cast iron 

Ditto upon brass 

Copper upon oak 

Yellow copper upon cast iron 

Leather (ox hide) well tanned upon cast 
iron, wetted 

Ditto upon brass, wetted 



Friction of 
Motion. 



"3.2 

am 



0-108 

143 
0-136 
0-119 
0-330 
0-140 
0-138 
0-177 

6 : i43 
0-160 
0-166 
0-107 

0-125 

0-137 
0-135 
0-144 
0-132 
0-107 
0-134 
0-100 
0-115 

0-229 
0-244 



.5 u & 
5,2-5 



6° 10 / 



18 16 



8 9 

9 6 

9 26 

6 7 

7 8 

7 49 

7 42 

8 12 



12 54 

13 43 



Friction op 
Quiescence. 



3§ 

S-43 
«.2 

3* 



0-390 
0-314 
6*420 

6 : il8 

6 : i66 

6*098 
6 : 164 



0-267 



s 

flog 
S.2-S 
.gel 



21° 19> 
17 26 

22**47 

'6**44 
*5**43 



9 19 



14 57 



342 



ELEMENTARY MECHANICS. 



TABLE in. 

EXPEDIENTS ON FRICTION WITH UNGUENTS INTBBPOSED. 

BY M. MORIN. 



The extent of the surfaces in these experiments bore such a relation to fcht 
pressure as to cause them to be separated from one another throughout by ar. 
interposed stratum of the unguent. 





Fbiction op 


Fbiction op 






Motion. 


Quiescence. 




SURFACES OF CONTACT. 






Unguents. 








Coefficient of 


Coefficient of 






Friction. 


Friction. 




Oak upon oak, fibres parallel 


0-164 


0-440 


Dry soap. 


Ditto ditto 


0*075 
0-067 
0-083 


0-164 
6-254 


Tallow. 


Ditto ditto 


Hog's lard. 
Tallow. 


Ditto, fibres perpendicular 


Ditto ditto 


0-072 
0-250 
0-136 




Hog's lard. 
Water. 


Ditto ditto 


Ditto upon elm, fibres parallel . . 


Dry soap. 


Ditto ditto 


0-073 
0-066 
0-080 


0-178 


Tallow. 


Ditto ditto 


Hog's lard. 
Tallow. 


Ditto upon cast iron, ditto 


Ditto upon wrought iron, ditto.. 


0-098 




TaUow. 


Beech upon oak, ditto 


0-055 
0-137 
0-070 
0-060 
0-139 


6 : 41*i 
0-142 

6*217 


Tallow. 


Elm upon oak, ditto 


Dry soap. 
Tallow. 


Ditto ditto 


Ditto ditto 


Hog's lard. 
Dry soap. 
Tallow. 


Ditto upon elm, ditto 


Ditto upon cast iron, ditto 


0-066 










( Greased ai d 


Wrought iron upon oak, ditto 


0-256 


0-649 


< saturated 
( with water 
Dry soap. 
Tallow. 


Ditto ditto ditto 


0-214 




Ditto ditto ditto 


0-085 


0-108 


Ditto upon elm, ditto 


0-078 




Tallow. 


Ditto ditto ditto 


0-076 




Hog's lard. 


Ditto ditto ditto 


0-055 




Olive oil. 


Dittc upon ;ast iron, ditto 


0-103 




Tallow. 


Ditto ditto ditto 


0-076 




Hog's lard. 


Ditto ditto ditto 


0-066 


0-100 


Olive oil 


Ditto upon wrought iron, ditto. . 


0-082 




Tallow. 


Ditto ditto ditto 


0-081 




Hog's lard. 


Ditto ditto ditto 


0-070 


0-115 


Olive oil. 


Wrought iron upon brass, fibres | 
parallel ) 








0-103 




Tallow. 


Ditto ditto ditto 


0-075 




Hog's lard 


Ditto ditto ditto 


0-078 




Olive oil. 


CJfcst iron upon oak, ditto 


0-189 




Dry soap. 





APPENDIX. 
TABLE III. — Continued. 



343 



SURFACES OF CONTACT. 



Cast iron upon oak, fibres parallel, 

Ditto ditto ditto 

Ditto ditto ditto 

Ditto ditto ditto 

Ditto upon elm, ditto , 

Ditto ditto ditto 

Ditto ditto ditto 

Ditto upon wrought iron , 

Cast iron upon cast iron 

Ditto ditto 

Ditto ditto 

Ditto ditto 

Ditto ditto 

Ditto ditto 

Ditto upon brass 

Ditto ditto 

Ditto ditto 

Copper upon oak, fibres parallel. . . 
Yellow copper upon cast iron 

Ditto ditto 

Ditto ditto 

Brass upon cast iron 

Ditto ditto 

Ditto upon wrought iron 

Ditto ditto , 

Ditto ditto 

Ditto upon brass 

Steel upon cast iron 

Ditto ditto 

Ditto ditto 

Ditto upon wrought iron 

Ditto ditto 

Ditto upon brass 

Ditto ditto , 

Ditto ditto 

Tanned ox hide upon cast iron 

Ditto ditto 

Ditto ditto ' 

Ditto upon brass 



Fbiction of Friction of 
Motion. ' Queesoence. 



Coefficient of 
Friction. 



0-218 



078 
075 
075 
077 
061 



0*091 



0-314 
0-197 
0-100 
0-070 
0-064 

0-055 

0-103 
0-075 

0-078 
0-069 
0-072 
0-068 
0-066 
0-086 
0-077 
0-081 



0-072 
0-058 
0-105 
0-081 
0-079 
0-093 
0-076 
0-056 
0-053 

0-067 



0-365 

0-159 
0-133 
0-241 



Coefficient of 
Friction. 



0*646 
0-100 

6 : i66 

0-100 



0-100 
0-100 



o-ioo 

0-103 



0-106 



0-108 



0.122 



Unguents. 



{Greased, and 
saturated 
with water. 
Tallow. 
Hog's lard. 
Olive oil. 
Tallow. 
Olive oil. 
j Hog's lard & 
\ plumbago. 
Tallow. 
Water. 
Soap. 
Tallow. 
Hog's lard. 
Olive oiL 
j Lard and 
i plumbago. 
Tallow. 
Hog's lard 
Olive oiL 
Tallow. 
Tallow. 
Hog's lard. 
Olive oil. 
Tallow. 
Olive oil 
Tallow. 
j Lard and 
( plumbago 
Olive oil. 
Olive oil 
Tallow. 
Hog's lard 
Olive oil. 
Tallow. 
Hog's lard 
TaUow. 
Olive oiL 
i Lard and 
] plumbago. 

(Greased, and 
saturated 
with water. 
Tallow. 
Olive oiL 
Tallow. 



344 



ELEMENTARY MECHANICS, 
TABLE III.— Continued. 



SURFACES OF CONTACT. 


Friction op 
Motion. 


Friction of 
Quiescence. 


UNOtrxiram. 


Coefficient of 
Friction. 


Coefficient of 
Friction. 


Tanned ox hide upon brass 

Ditto upon oak 

Hempen fibres not twisted, mov-' 
ing upon oak, the fibres of the 
hemp being placed in a direc- 
tion perpendicular to the direc- 
tion of the motion, and those of 
the oak parallel to it 


0-191 
0-29 

0-332 

0-194 
0-153 


6 : 79* 
0-869 

0-74 


Olive oiL 
Water. 

( Greased, and 
< saturated 
( with water. 


The same as above, moving upon 
cast iron 


Tallow. 


Ditto 


Olive oil 


Soft calcareous stone of Jaumont 
upon the same, with a layer of 
mortar, of sand, and lime inter- 
posed, after from 10 to 15 min- 
utes' contact 









TABLE IV. 

OF THE SPECIFIC GRAVITIES OF BODIES. 

[The density of distilled water is reckoned in this Table at its maximum, 
38%° F. = 1.000.] 



Name of the Body. 



Specific Gravity. 



L SOLID BODIES. 

(1) Metals. 

Antimony (of the laboratory) 

Brass 

Bronze for cannon, according to Lieut. Matzka 

Ditto, mean 

Copper, melted 

Ditto, hammered 

Ditto, wire-drawn 

Gold, melted 

Ditto, hammered 

Iron, wrought 



4-2 
7-6 

8-414 
8-758 
7-788 
8-878 
8-78 
19-238 
19-361 
7*207 



APPENDIX. 

TABLE IV .—Continued. 



345 



Name of the Body. 



L SOLID BODDilS. 

Iron, cast, a mean 

Ditto, gray 

Ditto, white 

Ditto for cannon, a mean 

Lead, pure melted 

Ditto, flattened 

Platinum, native 

Ditto, melted 

Ditto, hammered and wire-drawn 

Quicksilver, at 32° Fahr 

Silver, pure melted 

Ditto, hammered 

Steel, cast 

Ditto, wrought 

Ditto, much hardened 

Ditto, slightly 

Tin, chemically pure 

Ditto, hammered 

Ditto, Bohemian and Saxon 

Ditto, English 

Zinc, melted 

Ditto, rolled 

(2) Building Stones. 

Alabaster 

Basalt 

Dolerite 

Gneiss 

Granite 

Hornblende 

Limestone, various kinds 

Phonolite 

Porphyry 

Quartz 

Sandstone, various kinds, a mean 

Stones for building 

Syenite 

Trachyte 

Brick 

(3) Woods. 

Alder... 

Ash. 

Aspen 

Birch 

Box 

Elm. 

Fir 

Hornbeam 

Horse-chestnut 



Specific Gravity. 


7-251 


7*2 


7-5 


7-21 


11-3303 


11-388 


16-0 


20-855 


21-25 


13-568 


10-474 


10-51 


7-919 


7-840 


7-818 


7-833 


7-291 


7-299 


7-312 


7*291 


6-861 


7-191 


2-7 


2-8 


2*72 


2-5 


2*5 


2-9 


2-64 


2-51 


2-4 


2-56 


2-2 


1*66 


2-5 


2-4 


1-41 


Fresh felled. Dry. 


0-8571 0-5001 


0-9036 0-6440 


0-7654 0-4302 


. 0-9012 0-6274 


0-9822 0-5907 


0-9476 0-5474 


0-8941 0-5550 


0-9452 0-7695 


0-8614 0-5749 



15* 



346 



ELEMENTARY MECHANICS 
TABLE TV. — Continued. 



Name of the Body. 



Larch. 
Lime . 
Maple. 
Oak. . . 



I. SOLID BODIES. 



Ditto, another specimen. . 
Pine, Pinus Abies Picea , 
Ditto, Pinus Sylvestris. . 

Poplar (Italian) 

Willow , 

Ditto, white , 



(4) Vakious Solid Bodies. 

Charcoal, of cork , 

Ditto, soft wood , 

Ditto, oak 

Coal 



Coke , 

Earth, common 

rough sand 

rough earth, with gravel . , 

moist sand 

gravelly soil 

clay , 

clay or loam, with gravel 

Flint, dark. 

Ditto, white 

Gunpowder, loosely filled in, 

coarse powder 

musket ditto 

Ditto, slightly shaken down, 

musket powder , 

Ditto, solid 

Ice 

Lime, unslacked 

Resin, common , 

Rock salt 

Saltpetre, melted 

Ditto, crystallized 

Slate-pencil 

Sulphur 

Tallow 

Turpentine , 

Wax, white , 

Ditto, yellow 

Ditto, shoemaker's 



DL LIQUIDS. 

Acid, acetic 

Ditto, muriatic 

Ditto, nitric, concentrated 



Specific Gravity. 



Fresh felled. 
0-9206 
0-8170 
0-9036 
1-0494 
1-0754 
0-8699 
0-9121 
0*7634 
0-7155 
0-9859 



Dry. 
0-4735 
0-4390 
0-6593 
0-6777 
0-7075 
0-4716 
0-5502 
0-3931 
0-5289 
4873 



573 
232 

865 

48 

92 

02 

05 

07 

15 

48 

542 

741 



248 

916 

842 

089 

257 

745 

900 

8 

92 

942 

991 

969 

965 

897 



1-063 
1-211 
1-521 



APPENDIX. 
TABLE IV.— Continued. 



347 



Name of the Body. 



Specific Gravity. 



Acid, sulphuric, English 

Ditto, concentrated (Nordh.) 

Alcohol, free from water 

Ditto, common 

Ammoniac, liquid 

Aquafortis, double 

Ditto, single 

Beer 

Ether, acetic 

Ditto, muriatic 

Ditto, nitric 

Ditto, sulphuric 

Oil, linseed. 

Ditto, olive 

Ditto, turpentine 

Ditto, whale 

Quicksilver 

Water, distilled 

Ditto, rain 

Ditto, sea 

Wine 



III. GASES. 

Atmospheric air = tt a = 

Carbonic acid gas. 

Carbonic oxide gas , 

Carburetted hydrogen, a maximum 

Ditto from coals 

Chlorine 

Hydriodic gas 

Hydrogen 

Hydrosulphuric acid gas 

Muriatic acid gas 

Nitrogen 

Oxygen 

Phosphuretted hydrogen gas 

Steam at 212° Pahr 

Sulphurous acid gas 



1-845 
1-860 
0-792 
0-824 
0-875 
1-300 
1-200 
1-023 
0-866 
0-845 
0-886 
0-715 
0-928 
0-915 
0-792 
0-923 
13-568 
1-000 
1-0013 
1-0265 
0-992 



Barometer 30 In. 
Temperature =» 32* 
1-0000 
1-5240 
0-9569 
0-9784 
i 0-3000 
t 0-5596 
2-4700 
4-4430 
0-0688 
1-1912 
1-2474 
0-9760 
1-1026 
0-8700 
0-6235 
2-2470 



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Davis's Elevation and Stadia Tables 8vo, 1 00 

Folwell's Sewerage. (Designing and Maintenance.) 8vo, 3 00 

Freitag's Architectural Engineering. 2d Ed., Rewritten. . .8vo, 3 50 

Goodhue's Municipal Improvements 12mo, 1 75 

Goodrich's Economic Disposal of Towns' Refuse 8vo, 3 50 

Gore's Elements of Geodesy 8vo, 2 50 

Hayford's Text-book of Geodetic Astronomy 8vo, 3 00 

Howe's Retaining-walls for Earth 12mo, 1 25 

Johnson's Theory and Practice of Surveying Small 8vo, 4 00 

" Stadia and Earth-work Tables 8vo, 1 25 

Kiersted's Sewage Disposal 12mo, 1 25 

Mahan's Treatise on Civil Engineering. (1873.) (Wood.) . .8vo, 5 00 

* Mahan's Descriptive Geometry 8vo, 1 50 

Merriman's Elements of Precise Surveying and Geodesy. . . .8vo, 2 50 

Merriman and Brooks's Handbook for Surveyors. . . . 16mo, mor., 2 00 

Merriman's Elements of Sanitary Engineering 8vo, 2 00 

Nugent's Plane Surveying 8vo, 3 50 

Ogden's Sewer Design 12mo, 2 00 

Patton's Treatise on Civil Engineering 8vo, half leather, 7 50 

Reed's Topographical Drawing and Sketching 4to, 5 00 

Rideal's Sewage and the Bacterial Purification of Sewage. . . 8vo, 3 50 

Siebert and Biggin's Modern Stone-cutting and Masonry. . . .8vo, 1 50 

Smith's Manual of Topographical Drawing. (McMillan.) . .8vo, 2 50 

* Trautwine's Civil Engineer's Pocket-book. ... 16mo, morocco, 5 00 
Wait's Engineering and Architectural Jurisprudence 8vo, 6 00 

Sheep, 6 50 
" Law of Operations Preliminary to Construction in En- 
gineering and Architecture 8vo, 5 00 

Sheep, 5 50 
5 



Wait's Law of Contracts 8vo, 3 00 

Warren's Stereotomy — Problems in Stone-cutting 8vo, 2 50 

Webb's Problems in the Use and Adjustment of Engineering 

Instruments 16mo, morocco, 1 25 

* Wheeler's Elementary Course of Civil Engineering 8vo, 4 00 

Wilson's Topographic Surveying 8vo, 3 50 

BRIDGES AND ROOFS. 

Boiler's Practical Treatise on the Construction of Iron Highway 

Bridges 8vo, 2 00 

* Boiler's Thames River Bridge 4to, paper, 5 00 

Burr's Course on the Stresses in Bridges and Roof Trusses, 

Arched Ribs, and Suspension Bridges 8vo, . 3 50 

Du Bois's Mechanics of Engineering. Vol. II Small 4to, 10 00 

Foster's Treatise on Wooden Trestle Bridges 4to, 5 00 

Fowler's Coffer-dam Process for Piers 8vo, 2 50 

Greene's Roof Trusses 8vo, 1 25 

Bridge Trusses .8vo, 2 50 

" Arches in Wood, Iron, and Stone 8vo, 2 50 

Howe's Treatise on Arches 8vo, 4 00 

Johnson, Bryan and Turneaure's Theory and Practice in the 

Designing of Modern Framed Structures Small 4to, 10 00 

Merriman and Jacoby's Text-book on Roofs and Bridges: 

Part I.— Stresses in Simple Trusses , 8vo, 2 50 

Part IL-Graphic Statics . . . 8vo, 2 00 

Part III.— Bridge Design. Fourth Ed., Rewritten 8vo, 2 50 

Part IV.— Higher Structures 8vo, 2 50 

Morison's Memphis Bridge 4to, 10 00 

Waddell's De Pontibus, a Pocket Book for Bridge Engineers. 

1 61110, mor., 3 00 

" Specifications for Steel Bridges 12mo, 1 25 

Wood's Treatise on the Theory of the Construction of Bridges 

and Roofs 8vo, 2 00 

Wright's Designing of Draw-spans: 

Part I.— Plate-girder Draws 8vo, 2 50 

Part II. — Riveted-truss and Pin-connected Long-span Draws. 

8vo, 2 50 

Two parts in one volume 8vo, 3 50 



HYDRAULICS. 

Bazin's Experiments upon the Contraction of the Liquid Vein 

Issuing from an Orifice. (Trautwine.) 8vo, 2 00 

Bovey's Treatise on Hydraulics 8vo, 5 00 

Church's Mechanics of Engineering 8vo, 6 00 

Coffin's Graphical Solution of Hydraulic Problems. .16mo, mor., 2 50 

Flather's Dynamometers, and the Measurement of Power. 12mo, 3 00 

Folwell's Water-supply Engineering 8vo, 4 00 

Frizell's Water-power 8vo, 5 00 

Fuertes's Water and Public Health 12mo, 1 50 

" Water-filtration Works 12mo, 2 50 

Ganguillet and Kutter's General Formula for the Uniform 
Flow of Water in Rivers and Other Channels. (Her- 

ing and Trautwine.) 8vo, 4 00 

Hazen's Filtration of Public Water-supply 8vo, 3 00 

Hazlehurst's Towers and Tanks for Water-works 8vo, 2 50 

6 



Herschel's 115 Experiments on the Carrying Capacity of Large, 

Riveted, Metal Conduits 8vo, 2 00 

Mason's Water-supply. (Considered Principally from a Sani- 
tary Standpoint.) 8vo, 5 00 

Merriman's Treatise on Hydraulics 8vo, 4 00 

* Michie's Elements of Analytical Mechanics 8vo, 4 00 

Schuyler's Reservoirs for Irrigation, Water-power, and Domestic 

Water-supply Large 8vo, 5 00 

Turneaure and Russell. Public Water-supplies 8vo, 5 00 

Wegmann's Design and Construction of Dams 4to, 5 00 

" Water-supply of the City of New York from 1658 to 

1895 4to, 10 00 

Weisbach's Hydraulics and Hydraulic Motors. (Du Bois.) . .8vo, 5 00 

Wilson's Manual of Irrigation Engineering Small 8vo, 4 00 

Wolff's Windmill as a Prime Mover 8vo, 3 00 

Wood's Turbines 8vo, 2 50 

" Elements of Analytical Mechanics 8vo, 3 00 

MATERIALS OF ENGINEERING. 

Baker's Treatise on Masonry Construction 8vo, 5 00 

Black's United States Public Works Oblong 4to, 5 00 

Bovey's Strength of Materials and Theory of Structures 8vo, 7 50 

Burr's Elasticity and Resistance of the Materials of Engineer- 
ing ,..8vo, 5 00 

Byrne's Highway Construction 8vo, 5 60 

" Inspection of the Materials and Workmanship Em- 
ployed in Construction 16mo, 3 00 

Church's Mechanics of Engineering 8vo, 6 00 

Du Bois's Mechanics of Engineering. Vol. I Small 4to, 7 50 

Johnson's Materials of Construction Large 8vo, 6 00 

Keep's Cast Iron 8vo, 2 50 

Lanza's Applied Mechanics 8vo, 7 50 

Martens's Handbook on Testing Materials. (Henning.).2 v., 8vo, 7 50 

Merrill's Stones for Building and Decoration 8vo, 5 00 

Merriman's Text-book on the Mechanics of Materials 8vo, 4 00 

Merriman's Strength of Materials 12mo, 1 00 

Metcalf's Steel. A Manual for Steel-users 12mo, 2 00 

Patton's Practical Treatise on Foundations 8vo, 5 00 

Rockwell's Roads and Pavements in France 12mo, 1 25 

Smith's Wire: Its Use and Manufacture Small 4to, 3 00 

Snow's Properties Characterizing Economically Important 
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Spalding's Hydraulic Cement 12mo, 2 00 

Text-book on Roads and Pavements 12mo, 2 00 

Thurston's Materials of Engineering 3 Parts, 8vo, 8 00 

Part I. — Non-metallic Materials of Engineering and Metal- 
lurgy 8vo, 2 00 

Part II. — Iron and Steel 8vo, 3 50 

Part III. — A Treatise on Brasses, Bronzes and Other Alloys 

and Their Constituents 8vo, 2 50 

Thurston's Text-book of the Materials of Construction 8vo, 5 00 

Tillson's Street Pavements and Paving Materials 8vo, 4 00 

Waddell's De Pontibus. (A Pocket-book for Bridge Engineers.) 

16mo, morocco, 3 00 

" Specifications for Steel Bridges 12mo, 1 25 

Wood's Treatise on the Resistance of Materials, and an Ap- 
pendix on the Preservation of Timber 8vo, 2 00 

" Elements of Analytical Mechanics 8vo, 3 00 

7 



RAILWAY ENGINEERING. 

Andrews's Handbook for Street Railway Engineers. {In preparation.) 

Berg's Buildings and Structures of American Railroads ... 4to, 5 00 

Brooks's Handbook of Street Railroad Location. . 16mo, morocco, 1 50 

Butts's Civil Engineer's Field-book 16mo, morocco, 2 50 

Crandall's Transition Curve 16mo, morocco, 1 50 

Railway and Other Earthwork Tables 8vo, 1 50 

Dawson's Electric Railways and Tramways . Small 4to, half mor., 12 50 
" "Engineering" and Electric Traction Pocket-book. 

16mo, morocco, 4 00 

Dredge's History of the Pennsylvania Railroad: (1879.) .Paper, 5 00 

* Drinker's Tunneling, Explosive Compounds, and Rock Drills. 

4to, half morocco, 25 00 

Fisher's Table of Cubic Yards Cardboard, 25 

Godwin's Railroad Engineers' Field-book and Explorers' Guide. 

16mo, morocco, 2 50 

Howard's Transition Curve Field-book 16mo, morocco, 1 50 

Hudson's Tables for Calculating the Cubic Contents of Exca- 
vations and Embankments 8vo, 1 00 

Nagle's Field Manual for Railroad Engineers. . . .16mo, morocco, 3 00 

Philbrick's Field Manual for Engineers 16mo, morocco, 3 00 

Pratt and Alden's Street-railway Road-bed 8vo, 2 00 

Searles's Field Engineering 16mo, morocco, 3 00 

" Railroad Spiral 16mo, morocco, 1 50 

Taylor's Prismoidal Formulae and Earthwork 8vo, 1 50 

* Trautwine's Method of Calculating the Cubic Contents of Ex- 

cavations and Embankments by the Aid of Dia- 
grams 8vo, 2 00 

* " The Field Practice of Laying Out Circular Curves 

for Railroads 12mo, morocco, 2 50 

* Cross-section Sheet Paper, 25 

Webb's Railroad Construction 8vo, 4 00 

Wellington's Economic Theory of the Location of Railways. . 

Small 8vo, 5 00 



DRAWING. 

Barr's Kinematics of Machinery 8vo, 2 50 

* Bartlett's Mechanical Drawing 8vo, 3 00 

Durley's Elementary Text-book of the Kinematics of Machines. 

(In preparation.) 

Hill's Text-book on Shades and Shadows, and Perspective. . 8vo, 2 00 
Jones's Machine Design: 

Part I. — Kinematics of Machinery 8vo, 1 50 

Part II. — Form, Strength and Proportions of Parts 8vo, 3 00 

MacCord's Elements of Descriptive Geometry 8vo, 3 00 

Kinematics ; or, Practical Mechanism 8vo, 5 00 

Mechanical Drawing 4to, 4 00 

" Velocity Diagrams 8vo, I 50 

* Mahan's Descriptive Geometry and Stone-cutting 8vo, 1 50 

Mahan's Industrial Drawing. (Thompson.) 8vo, 3 50 

Reed's Topographical Drawing and Sketching 4to, 5 00 

Reid's Course in Mechanical Drawing 8vo, 2 00 

" Text-book of Mechanical Drawing and Elementary Ma- 
chine Design 8vo, 3 00 

Robinson's Principles of Mechanism 8vo, 3 00 



Smith's Manual of Topographical Drawing. (McMillan.) .8vo, 2 50 
Warren's Elements of Plane and Solid Free-hand Geometrical 

Drawing 12mo, 1 00 

" Drafting Instruments and Operations 12mo, 1 25 

" Manual of Elementary Projection Drawing 12mo, 1 50 

" Manual of Elementary Problems in the Linear Per- 
spective of Form and Shadow 12mo, 1 00 

" Plane Problems in Elementary Geometry 12mo, 1 25 

" Primary Geometry 12mo, 75 

" Elements of Descriptive Geometry, Shadows, and Per- 
spective 8vo, 3 50 

" General Problems of Shades and Shadows 8vo, 3 00 

" Elements of Machine Construction and Drawing. .8vo, 7 50 
" Problems, Theorems, and Examples in Descriptive 

Geometry 8vo, 2 50 

Weisbach's Kinematics and the Power of Transmission. (Herr- 
mann and Klein.) 8vo, 5 00 

Whelpley's Practical Instruction in the Art of Letter En- 
graving 12mo, 2 00 

Wilson's Topographic Surveying 8vo, 3 50 

Wilson's Free-hand Perspective 8vo, 2 50 

Woolf's Elementary Course in Descriptive Geometry. .Large 8vo, 3 00 



ELECTRICITY AND PHYSICS. 

Anthony and Brackett's Text-book of Physics. (Magie.) 

Small 8vo, 3 00 
Anthony's Lecture-notes on the Theory of Electrical Measur- 

ments 12mo, 1 00 

Benjamin's History of Electricity. 8vo, 3 00 

Benjamin's Voltaic Cell 8vo, 3 00 

Classen's Qantitative Chemical Analysis by Electrolysis. Her- 

rick and Boltwood.) 8vo, 3 00 

Crehore and Squier's Polarizing Photo-chronograph 8vo, 3 00 

Dawson's Electric Railways and Tramways.. Small 4to, half mor., 12 50 
Dawson's " Engineering " and Electric Traction Pocket-book. 

16mo, morocco, 4 00 

Mather's Dynamometers, and the Measurement of Power. .12mo, 3 00 

Gilbert's De Magnete. (Mottelay.) 8vo, 2 50 

Holman's Precision of Measurements 8vo, 2 00 

Telescopic Mirror-scale Method, Adjustments, and 

Tests Large 8vo, 75 

Landauer's Spectrum Analysis. (Tingle.) 8vo, 3 00 

Le Chatelier's High- temperature Measurements. (Boudouard — 

Burgess.) 12mo, 3 00 

L5b's Electrolysis and Electrosynthesis of Organic Compounds. 

(Lorenz.) 12mo, 1 00 

Lyons's Treatise on Electromagnetic Phenomena 8vo, 6 00 

* Michie. Elements of Wave Motion Relating to Sound and 

Light 8vo, 4 00 

Niaudet's Elementary Treatise on Electric Batteries (Fish- 
back.) 12mo, 2 50 

* Parshall and Hobart's Electric Generators-Small 4to, half mor., 10 00 
Ryan, Norris, and Hoxie's Electrical Machinery. (In preparation.) 
Thurston's Stationary Steam-engines 8vo, 2 50 

* Tillman. Elementary Lessons in Heat 8vo, 1 50 

Tory and Pitcher. Manual of Laboratory Physics .. Small 8vo, 2 00 

9 



LAW. 

* Davis. Elements of Law 8vo, 2 50 

Treatise on the Military Law of United States. .8vo, 7 00 

* Sheep, 7 50 

Manual for Courts-martial 16mo, morocco, 1 50 

Wait's Engineering and Architectural Jurisprudence 8vo, 6 00 

Sheep, 6 50 
" Law of Operations Preliminary to Construction in En- 
gineering and Architecture 8vo, 5 00 

Sheep, 5 50 

" Law of Contracts 8vo, 3 00 

Winthrop's Abridgment of Military Law 12mo, 2 5® 



MANUFACTURES. 

Beaumont's Woollen and Worsted Cloth Manufacture 12mo, 1 50 

Bernadou's Smokeless Powder — Nitro-cellulose and Theory of 

the Cellulose Molecule !2mo, 2 50 

Bolland's Iron Founder 12mo, cloth, 2 50 

" The Iron Founder " Supplement 12mo, 2 50 

" Encyclopedia of Founding and Dictionary of Foundry 

Terms Used in the Practice of Moulding. .. .12mo, 3 00 

Eissler's Modern Hi^h Explosives 8vo, 4 00 

Effront's Enzymes and their Applications. (Prescott.).. .8vo, 3 00 

Fitzgerald's Bost* i Machinist 18mo, 1 00 

Ford's Boiler Making for Boiler Makers 18mo, 1 00 

Hopkins's Oil-chemists' Handbook 8vo, 3 00 

Keep's Cast Iron 8vo 2 50 

Leach's The Inspection and Analysis of Food with Special 
Reference to State Control. {In preparation.) 

Metcalf's Steel. A Manual for Steel-users 12mo, 2 00 

Metcalf's Cost of Manufactures — And the Administration of 

Workshops, Public and Private 8vo, 5 00 

Meyer's Modern Locomotive Construction 4to, 10 00 

* Reisig's Guide to Piece-dyeing 8vo, 25 00 

Smith's Press- working of Metals 8vo, 3 00 

" Wire: Its Use and Manufacture Small 4to, 3 00 

Spalding's Hydraulic Cement 12mo, 2 00 

Spencer's Handbook for Chemists of Beet-sugar Houses. 

16mo, morocco, 3 00 
" Handbook for Sugar Manufacturers and their Chem- 
ists 16mo, morocco, 2 00 

Thurston's Manual of Steam-boilers, their Designs, Construc- 
tion and Operation 8vo, 5 00 

Walke's Lectures on Explosives 8vo, 4 00 

West's American Foundry Practice 12mo, 2 50 

" Moulder's Text-book 12mo, 2 50 

Wiechmann's Sugar Analysis Small 8vo, 2 50 

Wolff's Windmill as a Prime Mover 8vo, 3 00 

Woodbury's Fire Protection of Mills 8vo, 2 50 



MATHEMATICS. 

Baker's Elliptic Functions 8vo, 1 60 

* Bass's Elements of Differential Calculus 12mo, 4 00 

Briggs's Elements of Plane Analytic Geometry 12mo, 1 00 

10 



Chapman's Elementary Course in Theory of Equations. . .12mo, 

Compton's Manual of Logarithmic Computations 12mo, 

Davis's Introduction to the Logic of Algebra 8vo, 

De Laplace's Philosophical Essay on Probabilities. (Truscott 
and Emory.) (In preparation.) 

•Dickson's College Algebra Large 12mo, 

Halsted's Elements of Geometry 8vo, 

Elementary Synthetic Geometry 8vo, 

* Johnson's Three-place Logarithmic Tables : Vest-pocket size, 

pap., 

100 copies for 

Mounted on heavy cardboard, 8 X 10 inches, 

10 copies for 
Elementary Treatise on the Integral Calculus. 

Small 8vo, 

Curve Tracing in Cartesian Co-ordinates 12mo, 

Treatise on Ordinary and Partial Differential 

Equations Small 8vo, 

" Theory of Errors and the Method of Least 
Squares 12mo, 

* " Theoretical Mechanics w . . 12mo, 

* Ludlow and Bass. Elements of Trigonometry and Logarith- 

mic and Other Tables 8vo, 

Trigonometry. Tables published separately. .Each, 

Merriman and Woodward. Higher Mathematics 8vo, 

Merriman's Method of Least Squares 8vo, 

Rice and Johnson's Elementary Treatise on the Differential 

Calculus Small 8vo, 

" Differential and Integral Calculus. 2 vols. 

in one Small 8vo, 

Wood's Elements of Co-ordinate Geometry 8vo, 

" Trigometry: Analytical, Plane, and Spherical. ... 12mo, 



MECHANICAL ENGINEERING. 

MATERIALS OF ENGINEERING, STEAM ENGINES 
AND BOILERS. 

Baldwin's Steam Heating for Buildings 12mo, 2 50 

Barr's Kinematics of Machinery 8vo, 2 50 

* Bartlett's Mechanical Drawing 8vo, 3 00 

Benjamin's Wrinkles and Recipes 12mo, 2 00 

Carpenter's Experimental Engineering 8vo, 6 00 

Heating and Ventilating buildings 8vo, 3 00 

Clerk's Gas and Oil Engine Small 8vo, 4 00 

Cromwell's Treatise on Toothed Gearing 12mo, 1 50 

Treatise on Belts and Pulleys 12mo, 1 50 

Durley's Elementary Text-book of the Kinematics of Machines. 

(In preparation.) 

Flather's Dynamometers, and the Measurement of Power . . 12mo, 3 00 

" Rope Driving 12mo, 2 00 

Gill's Gas an Fuel Analysis for Engineers 12mo, 1 25 

Hall's Car Lubrication 12mo, 1 00 

Jones's Machine Design: 

Part I. — Kinematics of Machinery 8vo, 1 50 

Part II. — Form, Strength and Proportions of Parts 8vo, 3 00 

Kent's Mechanical Engineers' Pocket-book 16mo, morocco, 5 00 

Kerr's Power and Power Transmission 8vo, 2 00 

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MacCord's Kinematics; or, Practical Mechanism 8vo, 5 00 

Mechanical Drawing 4to, 4 00 

" Velocity Diagrams 8vo, 1 50 

Mahan's Industrial Drawing. (Thompson.) 8vo, 3 50 

Poole's Calorific Power of Fuels 8vo, 3 00 

Reid's Course in Mechanical Drawing 8vo, 2 00 

" Text-book of Mechanical Drawing and Elementary 

Machine Design 8vo, 3 00 

Richards's Compressed Air 12mo, 1 50 

Robinson's Principles of Mechanism 8vo, 3 00 

Smith's Press-working of Metals 8vo, 3 00 

Thurston's Treatise on Friction and Lost Work in Machin- 
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Animal as a Machine and Prime Motor and the 

Laws of Energetics 12mo, 1 00 

Warren's Elements of Machine Construction and Drawing. .8vo, 7 50 
Weisbach's Kinematics and the Power of Transmission. (Herr- 
mann—Klein.) 8vo, 5 00 

Machinery of Transmission and Governors. (Herr- 
mann—Klein.) 8vo, 5 00 

Hydraulics and Hydraulic Motors. (Du Bois.) .8vo, 5 00 

Wolff's Windmill as a Prime Mover 8vo, 3 00 

Wood's Turbines 8vo, 2 50 

MATERIALS OF ENGINEERING. 

Bovey's Strength of Materials and Theory of Structures .. 8vo, 7 50 
Burr's Elasticity and Resistance of the Materials of Engineer- 
ing 8vo, 5 00 

Church's Mechanics of Engineering 8vo, 6 00 

Johnson's Materials of Construction Large 8vo, 6 00 

Keep's Cast Iron 8vo, 2 50 

Lanza's Applied Mechanics 8vo, 7 50 

Martens's Handbook on Testing Materials. (Henning.) 8vo, 7 50 

Merriman's Text-book on the Mechanics of Materials .... 8vo, 4 00 

" Strength of Materials 12mo, 1 00 

Metcalf's Steel. A Manual for Steel-users 12mo, 2 00 

Smith's Wire: Its Use and Manufacture Small 4to, 3 00 

Thurston's Materials of Engineering. 3 vols., 8vo, 8 00 

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Part III. — A Treatise on Brasses, Bronzes and Other Alloys 

and their Constituents 8vo, 2 50 

Thurston's Text-book of the Materials of Construction .... 8vo, 5 00 
Wood's Treatise on the Resistance of Materials and an Ap- 
pendix on the Preservation of Timber .8vo, 2 00 

" Elements of Analytical Mechanics 8vo, 3 00 

STEAM ENGINES AND BOILERS. 

Carnot's Reflections on the Motive Power of Heat. (Thurston.) 

12mo, 1 50 
Dawson's " Engineering " and Electric Traction Pocket-book. 

16mo, morocco, 4 00 

Ford's Boiler Making for Boiler Makers 18mo, 1 00 

Goss's Locomotive Sparks 8vo, 2 00 

Hemenway's Indicator Practice and Steam-engine Economy. 

12mo, 2 00 

Hutton's Mechanical Engineering of Power Plants 8vo, 5 00 

" Heat and Heat-engines 8vo, 5 00 

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Kent's Steam-boiler Economy 8vo, 

Kneass's Practice and Theory of the Injector 8vo, 

MacCord's Slide-valves 8vo, 

Meyer's Modern Locomotive Construction 4to, 

Peabody's Manual of the Steam-engine Indicator 12mo, 

Tables of the Properties of Saturated Steam and 

Other Vapors 8vo, 

" Thermodynamics of the Steam-engine and Other 

Heat-engines 8vo, 

Valve-gears for Steam-engines 8vo, 

Peabody and Miller. Steam-boilers 8vo, 

Pray's Twenty Years with the Indicator Large 8vo, 

Pupin's Thermodynamics of Reversible Cycles in Gases and 

Saturated Vapors. (Osterberg.) 12mo, 

Reagan's Locomotive Mechanism and Engineering 12mo, 

Rontgen's Principles of Thermodynamics. (Du Bois.) . . . .8vo, 
Sinclair's Locomotive Engine Running and Management. .12mo, 
Smart's Handbook of Engineering Laboratory Practice. .12mo, 

Snow's Steam-boiler Practice 8vo, 

Spangler's Valve-gears 8vo, 

" Notes on Thermodynamics 12mo, 

Thurston's Handy Tables 8vo, 

" Manual of the Steam-engine 2 vols., 8vo, 

Part I. — History, Structure, and Theory 8vo, 

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Thurston's Handbook of Engine and Boiler Trials, and the Use 

of the Indicator and the Prony Brake 8vo, 

" Stationary Steam-engines 8vo, 

Steam-boiler Explosions in Theory and in Prac- 
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Manual of Steam-boilers, Their Designs, Construc- 
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Weisbach's Heat, Steam, and Steam-engines. (Du Bois.)..8vo, 

Whitham's Steam-engine Design 8 vo, 

Wilson's Treatise on Steam-boilers- (Flather.) 16mo, 

Wood's Thermodynamics, Heat Motors, and Refrigerating 

Machines 8vo, 4 00 



MECHANICS AND MACHINERY. 

Barr's Kinematics of Machinery 8vo, 

Bovey's Strength of Materials and Theory of Structures. .8vo, 

Chordal. — Extracts from Letters 12mo, 

Church's Mechanics of Engineering 8vo, 

" Notes and Examples in Mechanics 8vo, 

Compton's First Lessons in Metal-working 12mo, 

Compton and De Groodt. The ^pped Lathe 12mo, 

Cromwell's Treatise on Toothed Gearing 12mo, 

Treatise on Belts and Pulleys 12mo, 

Dana's Text-book of Elementary Mechanics for the Use of 

Colleges and Schools 12mo, 

Dingey's Machinery Pattern Making 12mo, 

Dredge's Record of the Transportation Exhibits Building of the 

World's Columbian Exposition of 1893 4to, half mor., 

Du Bois's Elementary Principles of Mechanics: 

Vol. I. — Kinematics 8vo, 

Vol. II.— Statics 8vo, 

Vol. III.— Kinetics 8vo, 

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Du Bois's Mechanics of Engineering. Vol. I Small 4to, 7 50 

Vol.11 Small 4to, 10 00 

Durley'g Elementary Text-book of the Kinematics of Machines. 

{In preparation.) 

Fitzgerald's Boston Machinist 16mo, 1 00 

Flather's Dynamometers, and the Measurement of Power. 12mo, 3 00 

" Rope Driving 12mo, 2 00 

Ooss's Locomotive Sparks 8vo, 2 00 

Hall's Car Lubrication 12mo, 1 00 

Holly's Art of Saw Filing 18mo, 75 

* Johnson's Theoretical Mechanics 12mo, 3 00 

Johnson's Short Course in Statics by Graphic and Algebraic 

Methods. (In preparation.) 
Jones's Machine Design: 

Part I. — Kinematics of Machinery 8vo, 1 50 

Part II. — Form, Strength and Proportions of Parts 8vo, 3 00 

Kerr's Power and Power Transmission 8vo, 2 00 

Lanza's Applied Mechanics 8vo, 7 50 

MacCord's Kinematics; or, Practical Mechanism 8vo, 5 00 

" Velocity Diagrams 8vo, 1 50 

Merriman's Text-book on the Mechanics of Materials 8vo, 4 00 

* Michie's Elements of Analytical Mechanics 8vo, 4 00 

Reagan's Locomotive Mechanism and Engineering 12mo, 2 00 

Reid's Course in Mechanical Drawing 8vo, 2 00 

" Text-book of Mechanical Drawing and Elementary 

Machine Design 8vo, 3 00 

Richards's Compressed Air 12mo, 1 50 

Robinson's Principles of Mechanism 8vo, 3 00 

Ryan, Norris, and Hoxie's Electrical Machinery. {In preparation.) 

Sinclair's Locomotive-engine Running and Management. .12mo, 2 00 

Smith's Press-working of Metals 8vo, 3 00 

Thurston's Treatise on Friction and Lost Work in Machin- 
ery and Mill Work 8vo, 3 00 

" Animal as a Machine and Prime Motor, and the 

Laws of Energetics 12mo, 1 00 

Warren's Elements of Machine Construction and Drawing. .8vo, 7 50 
Weisbach's Kinematics and the Power of Transmission. 

(Herrman — Klein.) 8vo, 5 00 

" Machinery of Transmission and Governors. (Herr- 

(man — Klein.) , 8vo, 5 00 

Wood's Elements of Analytical Mechanics 8vo, 3 00 

" Principles of Elementary Mechanics 12mo, 1 25 

" Turbines 8vo, 2 50 

The World's Columbian Exposition of 1893 4to, 1 00 

METALLURGY. 

Egleston's Metallurgy of Silver, Gold, and Mercury: 

Vol. I.-Silver 8vo, 7 50 

Vol. II.— Gold and Mercury 8vo, 7 50 

Keep's Cast Iron 8vo, 2 50 

Kunhardt's Practice of Ore Dressing in Lurope 8vo, 1 50 

Le Chatelier's High- temperature Measurements. (Boudouard — 

Burgess.) 12mo, 3 00 

Metcalf's Steel. A Manual for Steel-users 12mo, 2 00 

Thurston's Materials of Engineering. In Three Parts 8vo, 8 00 

Part II.— Iron and Steel 8vo, 3 50 

Part III. — A Treatise on Brasses, Bronzes and Other Alloys 

and Their Constituents 8vo, 2 50 

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1 00 


1 25 


3 50 


12 50 


1 OU 


4 00 


1 59 


1 00 


2 00 



MINERALOGY. 

Barringer's Description of Minerals of Commercial Value. 

Oblong, morocco, 

Boyd's Resources of Southwest Virginia 8vo, 

" Map of Southwest Virginia Pocket-book form, 

Brush's Manual of Determinative Mineralogy. (Penfield.) .8vo, 

Chester's Catalogue of Minerals 8vo, paper, 

Cloth, 

" Dictionary of the Names of Minerals 8vo, 

Dana's System of Mineralogy Large 8vo, half leather, 

" First Appendix to Dana's New " System of Mineralogy." 

Large 8vo, 

" Text-book of Mineralogy 8vo, 

" Minerals and How to Study Them 12mo, 

" Catalogue of American Localities of Minerals . Large 8vo, 

* Manual of Mineralogy and Petrography 12mo, 

Egleston's Catalogue of Minerals and Synonyms 8vo, 2 50 

Hussak's The Determination of Rock-forming Minerals. 

(Smith.) Small 8vo, 2 00 

* Penfield's Notes on Determinative Mineralogy and Record of 

Mineral Tests 8vo, paper, 50 

Rosenbusch's Microscopical Physiography of the Rock-making 

Minerals. (Idding's.) 8vo, 5 00 

•Tillman's Text-book of Important Minerals and Rocks.. 8vo, 2 00 
Williams's Manual of Lithology 8vo, 3 00 



MINING. 

Beard's Ventilation of Mines 12mo, 2 50 

Boyd's Resources of Southwest Virginia 8vo, 3 00 

" Map of Southwest Virginia Pocket-book form, 2 00 

•Drinker's Tunneling, Explosive Compounds, and Rock 

Drills 4to, half morocco, 25 00 

Eissler's Modern High Explosives 8vo, 4 00 

Goodyear's Coal-mines of the Western Coast of the United 

States 12mo, 2 50 

Ihlseng's Manual of Mining 8vo, 4 00 

Kunhardt's Practice of Ore Dressing in Europe 8vo, 1 50 

O'Driscoll's Notes on the Treatment of Gold Ores 8vo, 2 00 

Sawyer's Accidents in Mines 8vo, 7 00 

Walke's Lectures on Explosives 8vo, 4 00 

Wilson's Cyanide Processes 12mo, 1 50 

Wilson's Chlorination Process 12mo, 1 50 

Wilson's Hydraulic and Placer Mining 12mo, 2 00 

Wilson's Treatise on Practical and Theoretical Mine Ventila- 
tion 12mo, 1 25 



SANITARY SCIENCE. 

FolwelPs Sewerage. (Designing, Construction and Maintenance.) 

8vo, 3 00 

" Water-supply Engineering 8vo, 4 00 

Fuertes's Water and Public Health 12mo, 1 50 

" Water-filtration Works 12mo, 2 50 

15 



Gerhard's Guide to Sanitary House-inspection 16mo, 1 00 

Goodrich's Economical Disposal of Towns' Ref use ... Demy 8vo, 3 50 

Hazen's Filtration of Public Water-supplies 8vo, 3 00 

Kiersted's Sewage Disposal 12mo, 1 25 

Leach's The Inspection and Analysis of Food with Special 

Reference to State Control. (In preparation.) 
Mason's Water-supply. (Considered Principally from a San- 
itary Standpoint. 3d Edition, Rewritten 8vo, 4 00 

" Examination of Water. (Chemical and Bacterio- 
logical.) 12mo, 1 25 

Merriman's Elements of Sanitary Engineering 8vo, 2 00 

Nichols's Water-supply. (Considered Mainly from a Chemical 

and Sanitary Standpoint.) (1883.) 8vo, 2 50 

Ogden's Sewer Design 12mo, 2 00 

* Price's Handbook on Sanitation 12mo, 1 50 

Richards's Cost of Food. A Study in Dietaries 12mo, 1 00 

Richards and Woodman's Air, Water, and Food from a Sani- 
tary Standpoint 8vo, 2 00 

Richards's Cost of Living as Modified by Sanitary Science . 12mo, 1 00 

* Richards and Williams's The Dietary Computer 8vo, 1 50 

RideaPs Sewage and Bacterial Purification of Sewage 8vo, 3 50 

Turneaure and Russell's Public Water-supplies 8vo, 5 00 

Whipple's Microscopy of Drinking-water 8vo, 3 50 

Woodhull's Notes on Military Hygiene 16mo, 1 50 



MISCELLANEOUS. 

Barker's Deep-sea Soundings 8vo, 2 00 

Emmons's Geological Guide-book of the Rocky Mountain Ex- 
cursion of the International Congress of Geologists. 

Large 8vo, 1 50 

Ferrel's Popular Treatise on the Winds 8vo, 4 00 

Haines's American Railway Management 12mo, 2 50 

Mott's Composition, Digestibility, and Nutritive Value of Food. 

Mounted chart, 1 25 

" Fallacy of the Present Theory of Sound 16mo, 1 00 

Ricketts's History of Rensselaer Polytechnic Institute, 1824- 

1894 Small 8vo, 3 00 

Rotherham's Emphasised New Testament. . . . Large 8vo, 2 00 

" Critical Emphasised New Testament 12mo, 1 50 

Steel's Treatise on the Diseases of the Dog ,8vo, 3 50 

Totten's Important Question in Metrology 8vo, 2 50 

The World's Columbian Exposition of 1893 4to, 1 00 

Worcester and Atkinson. Small Hospitals, Establishment and 
Maintenance, and Suggestions for Hospital Architecture, 

with Plans for a Small Hospital 12mo, 1 25 



HEBREW AND CHALDEE TEXT-BOOKS. 

Green's Grammar of the Hebrew Language 8vo, 3 00 

" Elementary Hebrew Grammar 12mo, 1 25 

" Hebrew Chrestomathy 8vo, 2 00 

Gesenius's Hebrew and Chaldee Lexicon to the Old Testament 

Scriptures. (Tregelles.) Small 4to, half morocco, 5 00 

Letteris's Hebrew Bible .8vo, 2 25 

16 



SEP 23 i 

1 COPY DEL. TOCAT.DfV, 
SEP, 23 1902 



30 



